Question 13 Marks
- Drive the expression for electric field at a point on the equatorial line of an electric dipole.
- Depict the orientation of the dipole in (i) stable, (ii) unstable equilibrium in a uniform electric field.
Answer
-
Let the point ‘P’ be at a distance ‘r’ from the mid point of the dipole.
$E_{+q}=\frac{q}{4\pi\varepsilon_0(r^2+a^2)}$
$E_{-q}=\frac{q}{4\pi\varepsilon_0(r^2+a^2)}$
Both are equal and their directions are as shown in the figure. Hence net electric field
$\overrightarrow{E}=[-(E_{+q}+E_{-q})\cos\theta]\hat{p}$
$=-\frac{2qa}{4\pi\varepsilon_0(r^2+a^2)^{\frac{3}{2}}}\hat{p}$
- Stable equilibrium, $\theta=0^\circ$
Unstable equilibrium, $\theta=180^\circ$
View full question & answer→Question 23 Marks
A point charge of $2.0\ \mu\ C$ is at the centre of a cubic Gaussian surface $9.0 \ cm$ on edge. What is the net electric flux through the surface?
AnswerNet electric flux $(\phi_{\text{Net}})$ through the cubic surface is given by,
$\phi_{\text{Net}}=\frac{\text{q}}{\in_0}$
Where,
$\in_0 =$ permittivity of free space
$= 8.854 \times 10^{-12}\ N^{-1}\ C^2m^{-2}$
$q =$ Net charge contained inside the cube $= 2.0\ \mu C\ = 2 \times 10^{-6 }C$
$\therefore\phi_{\text{Net}}=\frac{2\times10^{-6}}{8.854 \times10^{-12}}$
$= 2.26 \times\ 10^5\ Nm^2C^{-1}$
The net electric flux through the surface is $2.26 \times 10^5\ Nm^2C^{-1}.$
View full question & answer→Question 33 Marks
Suppose that the particle in Exercise in $1.33$ is an electron projected with velocity $v_x= 2.0 \times 10^6 m s^{–1}.$ If $E$ between the plates separated by $0.5 \ cm$ is $9.1 \times 10^2N/C,$ where will the electron strike the upper plate? $(|e|=1.6 \times 10^{–19}C, m_e = 9.1 \times 10^{–31} \ kg.)$
AnswerVelocity of the particle, $V_x = 2 .0 \times 10^6\ m/s$
Separation of the two plates, $d = 0.5 \ cm = 0.005\ m$
Electric field between the two plates, $E = 9.1 \times 10^2 N/C$
Charge on an electron, $q = 1.6 \times10^{-19} C$
Mass of an electron, $m_{e }= 9 .1 \times 10^{-31} \ kg$
Let the electron strike the upper plate at the end of plate $L,$ when deflection is $s.$ Therefore,
$\text{s}=\frac{\text{qEL}^2}{2\text{mv}^2\text{x}}$
$\text{L}=\sqrt{\frac{2\text{dmv}^2\text{x}}{\text{qE}}}$
$=\sqrt{\frac{2\times0.005\times9.1\times10^{-31}\times(2.0\times10^6)^2}{1.6\times10^{-10}\times9.1\times10^2}}$
$=\sqrt{0.025\times10^{-2}}=\sqrt{2.5\times10^{-4}}$
$= 1.6 \times 10^{-2} m$
$= 1.6 \ cm$
Therefore, the electron will strike the upper plate after travelling $1.6 \ cm.$
View full question & answer→Question 43 Marks
Figure 1.33 shows tracks of three charged particles in a uniform electrostatic field. Give the signs of the three charges. Which particle has the highest charge to mass ratio?
AnswerOpposite charges attract each other and same charges repel each other. It can be observed that particles 1 and 2 both move towards the positively charged plate and repel away from the negatively charged plate. Hence, these two particles are negatively charged. It can also be observed that particle 3 moves towards the negatively charged plate and repels away from the positively charged plate. Hence, particle 3 is positively charged.
The charge to mass ratio (emf) is directly proportional to the displacement or amount of deflection for a given velocity. Since the deflection of particle 3 is the maximum, it has the highest charge to mass ratio.
View full question & answer→Question 53 Marks
A conducting sphere of radius $10 \ cm$ has an unknown charge. If the electric field $20 \ cm$ from the centre of the sphere is $1.5 \times 10^3 N/C$ and points radially inward, what is the net charge on the sphere?
AnswerGiven,
Radius of conducting sphere, $ r = 10 \ cm = 0. 1\ m$
Electric field , $E = 1.5\ x\ 10^{3 }N/C$ at distance, $d = 20 \ cm = 0. 2\ m$
As, $\text{E}=\frac{1}{4\pi\in_0}\frac{\text{q}}{\text{r}^2}$
$\Rightarrow \text{q = E.}\ 4\pi\epsilon_0.\text{r}^2$
$q = 1.5 \times 10^3 \times 4\pi \times 8.854 \times 10^{-12} \times (0.2)^2
q = 6.67 \times 10^{-9} C$
Here, since electric field is directed radially inward charge $q$ is negative.
Thus,
$q = - 6. 67 x 10^{-9}\ C$
$= -6.67\ nC.$
View full question & answer→Question 63 Marks
What is the force between two small charged spheres having charges of $2 \times 10^{-7}\ C$ and $3 \times 10^{-7}\ C$ placed $30\ cm$ apart in air?
AnswerRepulsive force of magnitude $6 \times 10^{-3} N$
Charge on the first sphere$, q_1 = 2 \times 10^{-7}\ C$
Charge on the second sphere$, q_2 = 3 \times 10^{-7}\ C$
Distance between the spheres$, r = 30 \ cm = 0.3\ m$
Electrostatic force between the spheres is given by the relation,
$\text{F}=\frac{\text{q}_1\text{q}_2}{4\pi\in_0\text{r}^2}$
Where, $\in_0 =$ Permittivity of free space
$\frac{1}{4\pi\in_0}=9\times10^9\text{ Nm}^2\text{C}^{2}$
$\text{F}=\frac{9\times10^9\times2\times10^{-7}\times3\times10^{-7}}{(0.3)^2}=6\times10^{-3}\text{N}$
Hence, force between the two small charged spheres is $6 \times 10^{-3}\ N.$ The charges are of same nature. Hence, force between them will be repulsive.
View full question & answer→Question 73 Marks
Two large, thin metal plates are parallel and close to each other. On their inner faces, the plates have surface charge densities of opposite signs and of magnitude $17.0 \times 10–22 C/m^2$ . What is $E:$
- In the outer region of the first plate,
- In the outer region of the second plate, and
- Between the plates?
AnswerGiven, Surface charge density, $\sigma = 17. 0\ x\ 10^{-22}\ C /m^2$
- To the left of the plates, electric fields are equal and opposite as plates are close to each other electric field is zero as surface charge density in outer side is zero.
- To the right of the plates, electric fields are equal and opposite as plates are close to each other electric field is zero.
- Electric fields between the plates are in same direction as total $E.F.$ on both sides of plate due to $\sigma$ surface charge density $= \frac{\sigma}{\epsilon_0}$ electric field of inner side of plate $=\frac{\sigma}{2\epsilon_0}$
and for both plate $E =\frac{\sigma}{2\epsilon_0}+\frac{\sigma}{2\epsilon_0}$
$\text{E}=\frac{\sigma}{\epsilon_0}=\sigma\times4\pi\times9\times10^9$
$E = 17.0 \times 10^{-22} \times 4 \times 3.14 \times 9 \times 10^9$
$E = 1921.7 \times 10^{-13}$
$= 1.92 \times 10^{-10} N/C.$ View full question & answer→Question 83 Marks
A hollow charged conductor has a tiny hole cut into its surface. Show that the electric field in the hole is $(\sigma /2ε_0)$ $\hat{\text{n}}$ , where $\hat{\text{n}}$ is the unit vector in the outward normal direction, and $\sigma$ is the surface charge density near the hole.
AnswerLet us take a charged conductor with the hole filled up, as shown by shaded portion in the figure.
We find with the application of Gaussian theorem that field inside is zero and just outside is $\frac{\sigma}{\epsilon_0}\hat{\text{n}}.$
This field can be viewed as the superposition of the field $E_2$ due to the filled up hole plus the field $E_1$ due to the rest of the charged conductor.
The two fields $(E_1$ and $E_2)$ must be equal and opposite as the field vanishes inside the conductor. Thus, $E_1 - E_2 = 0$
Now, the field outside the conductor is given by
$\text{E}_1+\text{E}_2=\frac{\sigma}{\epsilon_0}$
$\therefore \ 2\ \text{E}_1=\frac{\sigma}{\epsilon_0}$
$\Rightarrow\ \text{E}_1=\frac{\sigma}{2\epsilon_0}$
Therefore, field in the hole $($due to the rest of the conductor$)$ is given as:
$\text{E}_1=\frac{\sigma}{2\epsilon_0}\hat{\text{n}}$ ($\hat{\text{n}}\rightarrow$ unit vector in the outward normal direction$)$ View full question & answer→Question 93 Marks
Obtain the formula for the electric field due to a long thin wire of uniform linear charge density λ without using Gauss’s law.
[Hint: Use Coulomb’s law directly and evaluate the necessary integral.]
AnswerConsider a long thin wire of uniform linear charge density, $\lambda$.
To find: Formula for electric field due to this wire at any point P at a perpendicular distance PC= r from the wire.
Consider a small element of length dx of the wire with centre O, such that OC = x.
Charge on the element, q = $\lambda$. dx
So, electric intensity at P due to the element is given by,
$\text{dE}=\frac{1}{4\pi\epsilon_0}\frac{\lambda\text{dx}}{\text{OP}^2}=\frac{\lambda.\text{dx}}{4\pi\epsilon_0(\text{r}^2+\text{x}^2)}$
Now, $\text{d}\vec{\text{E}}$ can be resolved into two rectangular components, that is $\text{d}\vec{\text{E}}$ $\cos\theta$ in a perpendicular direction and $\text{d}\vec{\text{E}}$ $\sin\theta$ in a parallel direction.
The parallel component will be cancelled by the parallel component of the field due to charge on a similar element dx of wire on the other half.
The radial components get added.
Therefore,
Effective component of electric intensity due to the charge element, dE' = $\text{d}\vec{\text{E}}\cos\theta$
$\text{dE}' = \frac{\lambda.\text{dx}\cos\theta}{4\pi\epsilon_0(\text{r}^2+\text{x}^2)} \dots\dots(1)$
From $\triangle$ OCP, x = r tan $\theta$
$\therefore\text{dx}=\text{r}\sec^2\theta\ \text{d}\theta$
$\text{Now}, \text{r}^2+\text{x}^2=\text{r}^2+\text{r}^2\tan^2\theta=\text{r}^2(1+\tan^2\theta)$
From equation (1), we have
$\text{dE}'=\frac{\lambda\text{r}\sec^2\theta\ \text{d}\theta}{4\pi\epsilon\text{r}^2\sec^2\theta}\cos\theta$
$\Rightarrow\text{dE}' = \frac{\lambda}{4\pi\epsilon_0\text{r}}\cos\theta\ \text{d}\theta$
Since the wire has infinite length, it's ends A and B are infinite distances apart.
Therefore, $\theta$ varies from $-\frac{\pi}{2}\text{to}+\frac{\pi}{2}$
So, Electric Intensity at P due to the whole wire is given by,
$\text{E}'=\int_{-\frac{\pi}{2}}^{+\frac{\pi}{2}}\frac{\lambda}{4\pi\epsilon_0\text{r}}\cos\theta\ \text{d}\theta$
$=\frac{\lambda}{2\pi\epsilon_0\text{r}},$ is the required electric field intensity. View full question & answer→Question 103 Marks
Suppose the spheres $A$ and $B$ in Exercise $1.12$ have identical sizes. A third sphere of the same size but uncharged is brought in contact with the first, then brought in contact with the second, and finally removed from both. What is the new force of repulsion between $A$ and $B$?
AnswerDistance between the spheres, $A$ and $B, r = 0.5\ m$
Initially, the charge on each sphere, $q = 6.5\ x\ 10^{-7} C$
When sphere $A$ is touched with an uncharged sphere $C, \frac{\text{q}}{2}$ amount of charge from A will transfer to sphere $C.$ Hence, charge on each of the spheres, $A$ and $C,$ is $\frac{\text{q}}{2}$.
When sphere $C$ with charge $\frac{\text{q}}{2}$ is brought in contact with sphere $B$ with charge $q,$ total charges on the system will divide into two equal halves given as,
$\frac{\frac{\text{q}}{2}+\text{q}}{2}=\frac{3\text{q}}{4}$
Each sphere will share each half. Hence, charge on each of the spheres, $C$ and $B,$ is $\frac{3\text{q}}{4}.$
Force of repulsion between sphere A having charge $\frac{\text{q}}{2}$ and sphere $B$ having charge
$\frac{3\text{q}}{4}=\frac{\frac{\text{q}}{2}\times\frac{3\text{q}}{4}}{4\pi\in_0\text{r}^2}=\frac{3\text{q}^2}{8\times4\pi\in_0\text{r}^2}$
$=9\times10^9\times\frac{3\times(6.5\times 10^{-7})^2}{8\times(0.5)^2}$
$= 5.703 \times 10^{-3} N$
Therefore, the force of attraction between the two spheres is $5.703\ x\ 10^{-3} N.$
View full question & answer→Question 113 Marks
A point charge $+10\ \mu\ C$ is a distance $5\ cm$ directly above the centre of a square of side $10\ cm,$ as shown in Fig. $1.34.$ What is the magnitude of the electric flux through the square?
$($Hint: Think of the square as one face of a cube with edge $10\ cm.)$
AnswerThe square can be considered as one face of a cube of edge $10\ cm,$ with a centre where charge $q$ is placed. According to Gauss's theorem for a cube, total electric flux is through all its six faces.
$\phi_{\text{Total}}=\frac{\text{q}}{\in_0}$
Hence, electric flux through one face of the cube i.e., through the square, $\phi=\frac{\phi_{\text{Total}}}{6}$
$=\frac{1}{6}\frac{\text{q}}{\in_0}$
Where,
$\in_0 =$ Permittivity of free space
$= 8.854 \times 10^{-12} N^{-1}C^{-1}m^{-2}$
$q = 10\ \mu\ C = 10 \times 10^{-6 }C$
$\therefore\phi=\frac{1}{6}\times\frac{10\times10^{-6}}{8.854\times10^{-12}}$
$= 1.88 \times 10^5\ Nm^2C^{-1}$
Therefore, electric flux through the square is $1.88\ x\ 10^5\ Nm^2C^{-1}.$
View full question & answer→Question 123 Marks
- Consider an arbitrary electrostatic field configuration. A small test charge is placed at a null point (i.e., where E = 0) of the configuration. Show that the equilibrium of the test charge is necessarily unstable.
- Verify this result for the simple configuration of two charges of the same magnitude and sign placed a certain distance apart.
Answer
- Let the equilibrium of the test charge be stable. If a test charge is in equilibrium and displaced from its position in any direction, then it experiences a restorinq force towards a null point, where the electric field is zero. All the field lines near the null point are directed inwards towards the null point. There is a net inward flux of electric field through a closed surface around the null point. According to Gauss's law, the flux of electric field through a surface, which is not enclosing any charge, is zero. Hence, the equilibrium of the test charge can be stable.
- Two charges of same magnitude and same sign are placed at a certain distance. The mid-point of the joining line of the charges is the null point. When a test charged is displaced along the line, it experiences a restoring force. If it is displaced normal to the joining line, then the net force takes it away from the null point. Hence, the charge is unstable because stability of equilibrium requires restoring force in all directions.
View full question & answer→Question 133 Marks
- A conductor A with a cavity as shown in Fig. 1.36(a) is given a charge Q. Show that the entire charge must appear on the outer surface of the conductor.
- Another conductor B with charge q is inserted into the cavity keeping B insulated from A. Show that the total charge on the outside surface of A is Q + q [Fig. 1.36(b)].
- A sensitive instrument is to be shielded from the strong electrostatic fields in its environment. Suggest a possible way.
Answer
- Let us take a Gaussian surface which is lying completely within the conductor and enclosing the cavity. According to the Gaussian theorem, the charge enclosed by Gaussian surface must be zero as electric field vanishes everywhere inside a conductor. Thus, electric field vanishes inside the cavity. Therefore, charges which are supplied to the conductor reside on its outer surface.
- Let us take a Gaussian surface inside the conductor which is quite close to the cavity. According to the Gaussian theorem,
$\phi_\text{E}=\int\text{E.ds}=\frac{\text{total charge}}{\epsilon_0}$
As the electric field inside the conductor is zero, the total charge which is enclosed by the gaussian surface must be zero. This requires, a charge of -q units to be induced on the inner surface of the hollow conductor A. But an equal and opposite charge +q units must appear on the outer surface of conductor A, so that the total charge on the outer surface of A is Q + q.
- Use a metallic surface to enclose the sensitive instrument completely safe and intact. Because of electrostatic shielding, the electric field inside the metal surface vanishes to zero and all charge will reside on outer surface.
View full question & answer→Question 143 Marks
Which among the curves shown in Fig. 1.35 cannot possibly represent electrostatic field lines?
Answer
- Figure (a) cannot represent electrostatic field lines since electrostatic field lines start or end only at 90° to the surface of the conductor.
- Figure (b) too cannot represent electrostatic field lines as electrostatic field lines do not start from a negative charge. Electric field lines always traverse from a region of positive charge to a region of negative charge.
- Electrostatic field lines are represented by figure (c).
- Figure (d) cannot represent electrostatic field lines since no two such lines of force can intersect each other.
- As electrostatic field lines cannot form closed loop, therefore figure (d) also does not represent electrostatic field lines.
View full question & answer→Question 153 Marks
Careful measurement of the electric field at the surface of a black box indicates that the net outward flux through the surface of the box is $8.0 \times 103\ Nm^2 / C.$
- What is the net charge inside the box?
- If the net outward flux through the surface of the box were zero, could you conclude that there were no charges inside the box? Why or Why not?
Answer
- Net outward flux through the surface of the box, $(\phi) = 8.0\ x\ 10^3 Nm^2/C$
For a body containing net charge a, flux is given by the relation,
$\phi = \frac{\text{q}}{\in_0}$
$\in_0 =$ Permittivity of free space
$= 8.854 \times 10^{-12 }\times 8.0 \times 10^3$
$q = \in_0\phi$
$= 8.854 \times 10^{-12} \times 8.0 \times 10^3$
$= 7.08 \times 10^{-8}$
$= 0.07\ \mu\ C$
Therefore, the net charge inside the box is $0.07\ \mu\ C.$
- No
Net flux piercing out through a body depends on the net charge contained in the body. If net flux is zero, then it can be inferred that net charge inside the body is zero. The body may have equal amount of positive and negative charges. View full question & answer→Question 163 Marks
A polythene piece rubbed with wool is found to have a negative charge of $3 \times 10^{-7} C$.
- Estimate the number of electrons transferred $($from which to which?$)$
- Is there a transfer of mass from wool to polythene?
Answer
- When polythene is rubbed against wool, a number of electrons get transferred from wool to polythene. Hence, wool becomes positively charged and polythene becomes negatively charged.
Amount of charge on the polythene piece $, q = -3 \times 10^{-7} C$
Amount of charge on an electron $, e = -1.6 \times 10^{-19} C$
$$Number of electrons transferred from wool to polythene $= n$
$n$ can be calculated using the relation,
$q = ne$
$\text{n} = \frac{\text{q}}{\text{e}}$
$=\frac{-3\times10^{-7}}{-1.6\times10^{-19}}$
$= 1.87 \times 10^{12}$
Therefore, the number of electrons transferred from wool to polythene is $1.87 \times 10^{12}$.
- Yes.
There is a transfer of mass taking place.
This is because an electron has mass $, m_e = 9.1 \times 10^{-3} kg$
Total mass transferred to polythene from wool $, m = m_e \times n$
$= 9.1 \times 10^{-31} \times 1.85 \times 10^{12}$
$= 1.706 \times 10^{-18} kg$
Hence, a negligible amount of mass is transferred from wool to polythene. View full question & answer→Question 173 Marks
Consider a uniform electric field $E = 3 \times 10^3 î N/C$.
- What is the flux of this field through a square of $10 \ cm$ on a side whose plane is parallel to the $yz$ plane?
- What is the flux through the same square if the normal to its plane makes a $60^\circ $ angle with the $x-$ axis ?
Answer
- Electric field intensity, $\vec{\text{E}} = 3 \times 10^3 î N/C$
Magnitude of electric field intensity, $|\vec{\text{E}}| = 3 \times 10^3N/C$
Side of the square ,$ s = 10 \ cm= 0.1 m$
Area of the square,$ A = s^2 = 0.01 m^2$
The plane of the square is parallel to the $y-z$ plane.
Hence, angle between the unit vector normal to the plane and electric field, $\theta = 0^\circ $
Flux ($\phi$) through the plane is given by the relation,
$\phi = |\vec{\text{E}}|\text{A}\cos\theta$
$= 3 \times 10^{3 }\times 0.01 \times \cos0^{\circ}$
$= 30 Nm^2/C$
- Plane makes an angle of $60^\circ $ with the $x-$ axis.
- Hence, $e = 60^\circ $
Flux, $\phi = |\vec{\text{E}}|\text{A}\cos\theta$
$= 3 \times 10^{3 }\times 0.01 \times \cos60^{\circ}$
$= 30 \times \frac{1}{2} = 15 Nm^2/C$ View full question & answer→Question 183 Marks
The electrostatic force on a small sphere of charge $0.4 \mu C$ due to another small sphere of charge $–0.8 \mu C$ in air is $0.2 N.$
- What is the distance between the two spheres?
- What is the force on the second sphere due to the first?
Answer
- Electrostatic force on the first sphere, $F = 0.2 N$
- Charge on this sphere, $q_1 = 0.4 µC = 0.4 \times 10^{-6} C$
Charge on the second sphere$, q_2 = -0.8 µC = 0.8 \times 10^{-6} C$
Electrostatic force between the spheres is given by the relation,
$\text{F}=\frac{\text{q}_1\text{q}_2}{4\pi\in_0\text{r}^2}$ And $\frac{1}{4\pi\in_0}=9\times10^{9}\text{Nm}^2\text{C}^{-2}$
Where, $\in_0\ =$ Permittivity of free space
And, $\frac{1}{4\pi\in_0} = 9\times10^9\text{Nm}^{-2}\text{C}^{-2}$
$\text{r}^2=\frac{\text{q}_1\text{q}_2}{4\pi\in_0\text{F}}$
$= 144 \times 10^{-4}$
r = $\sqrt{144\times10^{-4}}=0.12\text{m}$
The distance between the two spheres is $0.12m.$
- Both the spheres attract each other with the same force. Therefore, the force on the second sphere due to the first is $0.2N.$
View full question & answer→Question 193 Marks
An electric dipole with dipole moment $4 \times 10^{-9} Cm$ is aligned at $30^\circ $ with the direction of a uniform electric field of magnitude $5 \times 10^4 NC^{-1}.$ Calculate the magnitude of the torque acting on the dipole.
AnswerElectric dipole moment $, p = 4 \times 10^{-9} Cm$
Angle made by p with a uniform electric field $, \theta = 30^\circ $
Electric field $, E = 5 \times 10^4 NC^{-1}.$
Torque acting on the dipole is given by the relation,
$Τ = pE \sin\theta $
$= 4 \times 10^{-9} \times 5 \times 10^{4 }\times \sin 30$
$= 20 \times 10^{-5} \times \frac{1}{2}$
$= 10^{-4} Nm$
Therefore, the magnitude of the torque acting on the dipole is $10^{-4} Nm$.
View full question & answer→Question 203 Marks
Consider the charges q, q, and –q placed at the vertices of an equilateral triangle, as shown in Fig. 1.7. What is the force on each charge?
AnswerThe forces acting on charge $q$ at A due to charges $q$ at and $-q$ at $C$ are $F _{12}$ along BA and $F _{13}$ along $AC$ respectively, as show1 in Fig. 1.7. By the parallelogram law, the total force $F _1$ on the charg $q$ at $A$ is given by
$F _1=F \hat{ r }_1$ where $\hat{ r }_1$ is a unit vector along BC.
The force of attraction or repulsion for each pair of charges has th same magnitude $F=\frac{q^2}{4 \pi \varepsilon_0 l^2}$
The total force $F _2$ on charge $q$ at B is thus $F _2=F \hat{ r }_2$, where $\hat{ r }_2$ is unit vector along $AC$.
Similarly the total force on charge $-q$ at $C$ is $F _3=\sqrt{3} F \hat{ n }$, where $\hat{ n }$ is the unit vector along the direction bisecting the $\angle BCA$.
It is interesting to see that the sum of the forces on the three charges is zero, i.e.,
$
F _1+ F _2+ F _3=0
$
The result is not at all surprising. It follows straight from the fact that Coulomb's law is consistent with Newton's third law. The proof is left to you as an exercise.
View full question & answer→Question 213 Marks
Consider three charges $q_1, q_2, q_3$ each equal to $q$ at the vertices of an equilateral triangle of side $l$. What is the force on a charge $Q$ (with the same sign as q) placed at the centroid of the triangle, as shown in Fig. 1.6?
AnswerIn the given equilateral triangle $ABC$ of sides of length $l$, if we draw a perpendicular $AD$ to the side $BC$,
$AD = AC \cos 30^{\circ}=(\sqrt{3} / 2) l$ and the distance $AO$ of the centroid $O$ from $A$ is $(2 / 3) AD =(1 / \sqrt{3})$. By symmatry $AO = BO = CO$.
Thus,
Force $F _1$ on $Q$ due to charge $q$ at $A =\frac{3}{4 \pi \varepsilon_0} \frac{Q q}{l^2}$ along $AO$
Force $F _2$ on $Q$ due to charge $q$ at $B =\frac{3}{4 \pi \varepsilon_0} \frac{Q q}{l^2}$ along BO
Force $F _3$ on $Q$ due to charge $q$ at $C =\frac{3}{4 \pi \varepsilon_0} \frac{Q q}{l^2}$ along $CO$
The resultant of forces $F _2$ and $F _3$ is $\frac{3}{4 \pi \varepsilon_0} \frac{Q q}{l^2}$ along $OA$, by the parallelogram law. Therefore, the total force on $Q=\frac{3}{4 \pi \varepsilon_0} \frac{Q q}{l^2}(\hat{ r }-\hat{ r })$ $=0$, where $\hat{ r }$ is the unit vector along $OA$.
It is clear also by symmetry that the three forces will sum to zero. Suppose that the resultant force was non-zero but in some direction. Consider what would happen if the system was rotated through $60^{\circ}$ about $O$.
View full question & answer→Question 223 Marks
A charged metallic sphere A is suspended by a nylon thread. Another charged metallic sphere B held by an insulating
handle is brought close to A such that the distance between their centres is 10 cm, as shown in Fig. 1.4(a). The resulting repulsion of A is noted (for example, by shining a beam of light and measuring the deflection of its shadow on a screen). Spheres A and B are touched by uncharged spheres C and D respectively, as shown in Fig. 1.4(b). C and D are then removed and B is brought closer to A to a distance of 5.0 cm between their centres, as shown in Fig. 1.4(c). What is the expected repulsion of A on the basis of Coulomb’s law? Spheres A and C and spheres B and D have identical sizes. Ignore the sizes of A and B in comparison to the separation between their centres.
AnswerLet the original charge on sphere A be $q$ and that on B be $q^{\prime}$. At a distance $r$ between their centres, the magnitude of the electrostatic force on each is given by
$
F=\frac{1}{4 \pi \varepsilon_0} \frac{q q^{\prime}}{r^2}
$
neglecting the sizes of spheres A and B in comparison to $r$. When an identical but uncharged sphere $C$ touches $A$, the charges redistribute on A and $C$ and, by symmetry, each sphere carries a charge q/2. Similarly, after D touches B, the redistributed charge on each is $q^{\prime} / 2$. Now, if the separation between A and B is halved, the magnitude of the electrostatic force on each is
$
F^{\prime}=\frac{1}{4 \pi \varepsilon_0} \frac{(q / 2)\left(q^{\prime} / 2\right)}{(r / 2)^2}=\frac{1}{4 \pi \varepsilon_0} \frac{\left(q q^{\prime}\right)}{r^2}=F
$
Thus the electrostatic force on A, due to B, remains unaltered.
View full question & answer→Question 233 Marks
The electric field components in Fig. $1.24$ are $E_x=\alpha x^{1 / 2}, E_y=E_z=0$, in which $\alpha=800 N / C m ^{1 / 2}$. Calculate $(a)$ the flux through the cube, and $(b)$ the charge within the cube. Assume that $a=0.1 m$.
Answer$(a)$ Since the electric field has only an $x$ component, for faces perpendicular to $x$ direction, the angle between $E$ and $\Delta S$ is $\pm \pi / 2$.
Therefore, the flux $\phi= E . \Delta S$ is separately zero for each face of the cube except the two shaded ones.
Now the magnitude of the electric field at the left face is
$E_L=\alpha x^{1 / 2}=\alpha a^{1 / 2}$
$( x=a$ at the left face$).$
The magnitude of electric field at the right face is
$E_R=\alpha x^{1 / 2}=\alpha(2 a)^{1 / 2}$
$(x=2 a$ at the right face).
The corresponding fluxes are
$\phi_L = E _L \cdot \Delta S =\Delta S E _L \cdot \hat{ n }_L=E_L \Delta S \cos \theta=-E_L \Delta S ,$
since $\theta=180^{\circ}$
$ =-E_L a^2$
$\phi_R = E _R \cdot \Delta S =E_R \Delta S \cos \theta=E_R \Delta S ,$
since $\theta=0^{\circ}$
$ =E_R a^2$
Net flux through the cube
$=\phi_R+\phi_L=E_R a^2-E_L a^2=a^2\left(E_R-E_L\right)$
$=\alpha a^2\left[(2 a)^{1 / 2}-a^{1 / 2}\right]$
$=\alpha a^{5 / 2}(\sqrt{2}-1)$
$=800(0.1)^{5 / 2}(\sqrt{2}-1)$
$=1.05 N m ^2 C ^{-1}$
$(b)$ We can use Gauss's law to find the total charge $q$ inside the cube.
We have $\phi=q / \varepsilon_0$ or $q=\phi \varepsilon_0$.
Therefore $, q=1.05 \times 8.854 \times 10^{-12} C $
$=9.27 \times 10^{-12} C \text {. }$
View full question & answer→Question 243 Marks
If $10^9$ electrons move out of a body to another body every second, how much time is required to get a total charge of $1 C$ on the other body?
AnswerIn one second $10^9$ electrons move out of the body. Therefore the charge given out in one second is $1.6 \times 10^{-19} \times 10^9 C =1.6 \times 10^{-10} C$. The time required to accumulate a charge of $1 C$ can then be estimated to be $1 C \div\left(1.6 \times 10^{-10} C / s \right)=6.25 \times 10^9 s =6.25 \times 10^9 \div(365 \times 24 \times$ 3600) years $=198$ years. Thus to collect a charge of one coulomb, from a body from which $10^9$ electrons move out every second, we will need approximately 200 years. One coulomb is, therefore, a very large unit for many practical purposes.
It is, however, also important to know what is roughly the number of electrons contained in a piece of one cubic centimetre of a material. A cubic piece of copper of side $1 cm$ contains about $2.5 \times 10^{24}$ electrons.
View full question & answer→Question 253 Marks
A positive point charge (+ q) is kept in the vicinity of an uncharged conducting plate. Sketch electric field lines originating from the point on to the surface of the plate.
Derive the expression for the electric field at the surface of a charged conductor.
Answer
Field at the surface of a charged conductor 
We have, by Gauss’s law,$\text{E }\delta\text{ S} =\frac{|\sigma| \delta\text{ S}}{\varepsilon_{0}}$
$\therefore\text{E} = \frac{|\sigma|}{\varepsilon_{0}}.$ View full question & answer→Question 263 Marks
State Gauss's theorem in electrostatics. Apply this theorem to derive an expression for electric field intensity at a point near an infinitely long straight charged wire.
Answer
- Statement: Net electric flux through to a closed surface is equal to $\frac{1}{\varepsilon_\circ}$ times the total net charge enclosed within the surface.
(If the student just writes $\oint\text{E}.\text{ds} =\frac{\text{q}}{\in_\circ}$, award )
- Diagram:-
- Derivation:-
$\oint\text{E}.\text{ds} \int\limits_{s_1}\overline{E}.\text{d}\overline{s}_{1} + \int\limits_{s_2}\overline{E}.\text{ds}_{2} + \int\limits_{s_3}\overline{E}.\text{ds}_{3}$
$ = 0 + 0 + 2\pi\text{r}\ell$
Also, $\text{q} = \lambda$ $\ell$ (where $\lambda$ is charge per unit length)
$(\text{E}).(2\pi\text{r}\ell) = \frac{1}{\varepsilon_\circ}\lambda\ell $ OR $\text{E}2\pi\text{r}\ell\frac{\text{q}}{\varepsilon_\circ}$
$\text{E} = \frac{\lambda}{2\pi\varepsilon_\circ\text{r}}$ OR $\text{E} = \frac{\text{q}}{2\pi\varepsilon_\circ\text{r}\ell}$. View full question & answer→Question 273 Marks
Using Gauss’s theorem, show mathematically that for any point outside the shell, the field due to a uniformly charged thin spherical shell is the same as if the entire charge of the shell is concentrated at the centre. Why do you expect the electric field inside the shell to be zero according to this theorem?
Answer$\phi =\oint\limits_{s}\overrightarrow{\text{E}}.\overrightarrow{\text{d}}\text{s} = \frac{\text{q}}{\varepsilon_{\circ}}$
Derivation: $\text{E}\times4\pi\text{r}^{2} =\frac{\sigma}{\varepsilon_{\circ}}4\pi\text{R}^{2}$
$\therefore \text{E} = \frac{\sigma\text{R}^{2}}{\varepsilon_{\circ}\text{r}^{2}}$
where $\text{q} = 4\pi\text{R}^{2}\sigma$ is the total charge on the spherical shell. Electrostatic field is zero, since total charge inside the shell is zero or charge reside on the surface of the shell. View full question & answer→Question 283 Marks
Using Gauss’s law in electrostatics, deduce an expression for electric field intensity due to a uniformly charged infinite plane sheet. If another identical sheet is placed parallel to it, show that there is no electric field in the region between the two sheets.
Answer
By Gauss’s law $ \oint \overrightarrow{E}.\overrightarrow{ds}=\frac{q}{\epsilon_0}$
$\therefore{2{\text{EA}}}=\frac{\sigma{\text{A}}}{{\epsilon_0}}$
$\therefore\text{E}=\frac{\sigma}{2\epsilon_0}\text{ }\text{or}\frac{\sigma}{2\epsilon_0}\text{A}$
Electric field between two identical charged sheets 
$\because$ Both the sheets have same charge density, their electric fields will be equal and opposite in the region between the two sheets.
Hence the net field is zero.
Alternate Answer
$\text{E}_1=\frac{\sigma}{2\epsilon_0}$
$\text{E}_2=-\frac{\sigma}{2\epsilon_0}$
Resultant electric field between the plates $ = E_1 + E_2\frac{\sigma}{2\epsilon_0}-\frac{\sigma}{2\epsilon_0}=0$ View full question & answer→Question 293 Marks
Define electric flux. Write its SI unit.
Using Gauss’s law, deduce an expression for electric field intensity due to an infinitely long straight uniformly charged wire.
AnswerThe electric flux is defined as$\phi_E=\overrightarrow{E}.\overrightarrow{A}={EA}\cos\theta$
Its S.I unit is $(Nm^{2}C^{-1})$ The Gaussian surface is cylindrical and field is radial. At the cylindrical part of the surface, cylindrical part of the surface, $ \overrightarrow{E}$ is normal to the surface at every point and its magnitude is constant (since it depends only on r). By Gauss’s theorem : $ \oint \overrightarrow{E}.{d}\overrightarrow{S}=\frac{q}{\epsilon_0}$$\therefore(2\pi{rl})=\frac{\lambda{l}}{\epsilon_0}$
$\text{or}\text{ }\text{ } E=\frac{\lambda}{2\pi\epsilon_0r}$
View full question & answer→Question 303 Marks
- Define torque acting on a dipole of dipole moment $\overrightarrow{p}$ placed in a uniform electric field $\overrightarrow{\text{E}}$. Express it in the vector form and point out the direction along which it acts.
- What happens if the field is non-uniform?
- What would happen if the external field $\overrightarrow{\text{E}}$ is increasing (i) parallel to $\overrightarrow{p}$ and (ii) anti-parallel to $\overrightarrow{p}$?
Answer
- $\tau=pE\sin\theta\text{ };\text{ }\theta=$ angle between dipole moment($\overrightarrow{p}$) and electric field($\overrightarrow{\text{E}}$)
$\tau=\overrightarrow{p}\times\overrightarrow{\text{E}}$
Direction of torque is perpendicular to the plane containing and given by right-hand screw rule.
Alternate Answer
Direction of torque is out of the plane of the paper.
- If the field is non uniform the net force on the dipole will not be zero. There will be translatory motion of the dipole.
-
- Net force will be in the direction of increasing electric field.
- Net force will be in the direction opposite to the increasing field. [or in the direction of decreasing field]
View full question & answer→Question 313 Marks
A charge Q is distributed uniformly over a metallic sphere of radius R. Obtain the expressions for the electric field (E) and electric potential (V) at a point 0 < x < R.
Show on a plot the variation of E and V with x for 0 < x < 2R.
Answer
By Gauss theorem $\oint \overrightarrow{E}.d\vec{s}=\frac{q}{E_0}$ q = 0 in interval 0 < x < R$\Rightarrow E=0$
$E=-\frac{dv}{dr}$
$\Rightarrow V= constant=\frac{1}{4\pi E_0}\frac{Q}{R}$
View full question & answer→Question 323 Marks
A long charged cylinder of linear charge density $+\lambda_1$ is surrounded by a hollow coaxial conducting cylinder of linear charge density $-\lambda_2$. Use Gauss’s law to obtain expressions for the electric field at a point (i) in the space between the cylinders, and (ii) outside the larger cylinder.
AnswerAs Gauss’s Law states:$\oint\overrightarrow{E}.\vec{ds}=\frac{q}{\epsilon_0}$
- $\oint\overrightarrow{E_1}.\vec{ds}=\frac{\lambda_1l}{\epsilon_0}$
$\Longrightarrow\overrightarrow{E_1}=\frac{\lambda_1}{2\pi\epsilon_0r_1}\hat{r_1}$
- $\oint\overrightarrow{E_2}.\vec{ds}=\frac{(\lambda_1-\lambda_2)l}{\epsilon_0}$
$\Longrightarrow\overrightarrow{E_2}=\frac{(\lambda_1-\lambda_2)}{2\pi\epsilon_0r_2}\hat{r_2}$ View full question & answer→Question 333 Marks
- Define torque acting on a dipole of dipole moment $\overrightarrow{p}$ placed in a uniform electric field $\overrightarrow{\text{E}}$. Express it in the vector form and point out the direction along which it acts
- What happens if the field is non-uniform?
- What would happen if the external field $\overrightarrow{\text{E}}$ is increasing (i) parallel to $\overrightarrow{p}$ and (ii) anti-parallel to $\overrightarrow{p}$?
Answer
- $\tau=pE\sin\theta\text{ };\text{ }\theta=$ angle between dipole moment($\overrightarrow{p}$) and electric field($\overrightarrow{\text{E}}$)
$\tau=\overrightarrow{p}\times\overrightarrow{\text{E}}$
Direction of torque is perpendicular to the plane containing and given by right-hand screw rule.
Alternate Answer
Direction of torque is out of the plane of the paper.
- If the field is non uniform the net force on the dipole will not be zero. There will be translatory motion of the dipole.
-
- Net force will be in the direction of increasing electric field.
- Net force will be in the direction opposite to the increasing field. [or in the direction of decreasing field]
View full question & answer→Question 343 Marks
- Obtain the expression for the torque $\overrightarrow\tau$ experienced by an electric dipole of dipole moment $\overrightarrow{\text{P}}$ in a uniform electric field$\overrightarrow{\text{E}}$
- What will happen if the field were not uniform?
Answer
-
Force on + q, $\overrightarrow{\text{F}}=q\overrightarrow{\text{E}}$
Force on - q, $\overrightarrow{\text{F}}=-q\overrightarrow{\text{E}}$
Magnitude of torque
$\tau=q\text{E}\times2a \text{ sin}\theta$
$\tau=2q\text{a}\times{\text{E}} \text{ sin}\theta$
$\overrightarrow\tau=\overrightarrow{\text{P}}\times\overrightarrow{\text{E}}$
- If the electric field is non uniform, the dipole experiences a translatory force as well as a torque.
View full question & answer→Question 353 Marks
A charge is distributed uniformly over a ring of radius ‘a’. Obtain an expression for the electric intensity E at a point on the axis of the ring. Hence show that for points at large distances from the ring, it behaves like a point charge.
Answer
Net Electric Field at point $\text{P} =\int^{2\pi\text{a}}_{\circ}\text{dE}\cos\theta$
dE = Electric field due to a small element having charge dq
$ = \frac{1}{4\pi\varepsilon_\circ}\frac{\text{dq}}{\text{r}^{2}}$
Let $\lambda$ = Linear charge density
$ = \frac{\text{dq}}{\text{dl}}$
$\text{dq} = \lambda\text{dl}$
Hence $\text{E} = \int^{2\pi\text{a}}_{\circ}\frac{1}{4\pi\varepsilon_\circ}.\frac{\lambda\text{dl}}{\text{r}^{2}}\times\frac{x}{r} , $
where $\cos\theta =\frac{\text{x}}{\text{r}}$
$ = \frac{\lambda\text{x}}{4\pi\varepsilon_\circ\text{r}^{3}}(2\pi\text{a})$
$ =\frac{1}{4\pi\varepsilon_\circ}\frac{\text{Qx}}{(\text{x}^{2} + \text{a}^{2})^{\frac{3}{2}}}$
where total charge $\text{Q} = \lambda\times2\pi\text{a}$
At large distance i.e. x >> a
$\text{E}\simeq\frac{1}{4\pi\varepsilon_\circ}.\frac{\text{Q}}{\text{x}^{2}}$
This is the Electric field due to a point charge at distance x. View full question & answer→Question 363 Marks
- Derive the expression for the capacitance of a parallel plate area $A$ and plate separation $d$.
- Two charged spherical conductors of radii $R_1$ and $R_2$ when conducting wire acquire charges $q_1$ and $q_2$ respectively. surface charge densities in terms of their radii.
Answer
Electric field between the plates of capacitor $E = \frac{\sigma}{\varepsilon_{0}} =\frac{\text{Q}}{\text{A}\varepsilon_{0}}$
$\therefore$ potential difference
$\text{V} = \text{Ed} = \frac{\text{Qd}}{\text{A}\varepsilon_{0}}$
Capacitance
$\text{C} = \frac{\text{Q}}{\text{V}} = \frac{\varepsilon_{0}\text{A}}{\text{d}}$
- When the two charged spherical conductors are connected by a conducting wire ,
- they acquire the same potential
i.e $\frac{\text{Kq}_{1}}{\text{R}_{1}} = \frac{\text{Kq}_{2}}{\text{R}_{2}}$
$\Rightarrow\frac{\text{q}_{1}}{\text{q}_{2}} = \frac{\text{R}_{1}}{\text{R}_{2}}$
Hence, ratio of surface charge densities
$\frac{\sigma_{1}}{\sigma_{2}} = \frac{\text{q}_{1}/4\pi\text{R}_{1}^{2}}{\text{q}_{2}/4\pi\text{R}_{2}^{2}}$
$ = \frac{\text{q}_{1}\text{R}_{2}^{2}}{\text{q}_{2}\text{R}_{1}^{2}}$
$ = \frac{\text{R}_{1}}{\text{R}_{2}}\times\frac{\text{R}_{2}^{2}}{\text{R}_{1}^{2}} = \frac{\text{R}_{2}}{\text{R}_{1}}.$ View full question & answer→Question 373 Marks
A capacitor of unknown capacitance is connected across a battery of $V$ volts. The charge stored in it is $360 \mu C$. When potential across the capacitor is reduced by $120 V,$ the charge stored in it becomes $120 \mu C$.
Calculate:
- The potential $V$ and the unknown capacitance $C$.
- What will be the charge stored in the capacitor, if the voltage applied had increased by $120 V$?
Answer
- Initial voltage $, V_1 = V$ volts and charge stored $, Q_1 = 360 µC$.
$Q_1 = CV_1 …(1)$
Changed potential $, V_2 = V − 120$
$Q_2 = 120 µC$
$Q_2 = CV_2 ...(2)$
By dividing $(2)$ from $(1),$
- we get $\frac{\text{Q}_{1}}{\text{Q}_{2}}=\frac{\text{C}{V}_{1}}{\text{C}{V}_{2}}$
- $\Rightarrow \frac{360}{120}=\frac{\text{V}}{\text{V}-{120}}$
$\therefore {\text{C}}=\frac{\text{Q}_{1}}{\text{V}_{1}}=\frac{{360}\times{10}{^-}{^6}}{180}={2}\times{10}{^-}{^6} \text{F}={2}\mu{\text{F}}$
- If the voltage applied had increased by $120 V,$ then $\text{V}_{3}={180}+{120}={300}\text{V}.$
Hence, charge stored in the capacitor, $\text{Q}_{3}=\text{C}\text{V}_{3}={2}\times{10}{^-}{^6}\times{300}={600}\mu\text{C}.$ View full question & answer→Question 383 Marks
A hollow cylindrical box of length $1 m$ and area of cross$-$section $25 \ cm^2$ is placed in a three dimensional coordinate system as shown in the figure. The electric field in the region is given by$\overrightarrow{\text{E}} = 50\text{x}\hat{\text{i}},$ where $E$ is in $NC^{–1}$ and $x$ is in metres. Find
- Net flux through the cylinder.
- Charge enclosed by the cylinder.
Answer
-
Given, $\vec{E}={50}{x}\vec{i}$ and $\bigtriangleup{s}={25}\text{ C}\text{m}{^2}={25}\times{10}{^-}{^4}\text{m}{^2}$
As the electric field is only along the $x-$axis, so, flux will pass only through the cross$-$section of cylinder.
Magnitude of Electric Field at cross $-$ section $A, \text{E}_\text{A}={50}\times{1}={50}\text{ N}\text{C}{^-}{^1}$
Magnitude of Electric Field at cross $-$ Section $B, \text{E}_\text{B}={50}\times{2}={100}\text{ N}\text{C}{^-}{^1}$
The Corresponding Electric Fluxes are:
$\phi_\text{A}=\vec{\text{E}}.\bigtriangleup\vec{s}={50}\times{25}\times{10}{^-}{^4}\times\cos{180}{^0}=-{0.125}\text{ N}\text{ m}{^2}\text{C}{^-}{^1}$
$\phi_\text{B}=\vec{\text{E}}.\bigtriangleup\vec{s}={100}\times{25}\times{10}{^-}{^4}\times\cos{0}{^0}=-{0.25}\text{ N}\text{ m}{^2}\text{C}{^-}{^1}$
So, the net flux through the cylinder, $\phi=\phi_\text{A}+\phi_\text{B}= -{0.125} + {0.25}={0.125}\text{ N}\text{ m}{^2}\text{C}{^-}{^1}$
- Using Gauss’s law:
$\oint\vec{\text{E}}{d}\vec{\text{s}}=\frac{\text{q}}{\epsilon_{0}}\Rightarrow{0.125}=\frac{\text{q}}{{8.85}\times{10}{^-}{^1}{^2}}$
$\Rightarrow\text{q}={8.85}\times{0.125}\times{10}{^-}{^1}{^2}={1.1}\times{10}{^-}{^1}{^2}\text{C}$ View full question & answer→Question 393 Marks
Using Gauss's law obtain the expression for the electric field due to a uniformly charged thin spherical shell of radius R at a point outside the shell. Draw a graph showing the variation of electric field with r, for r > R and r < R.
Answer
From Gauss's theorem, ${\phi = \oint\overrightarrow{\text{E}}}.\text{d}\overrightarrow{\text{S}} = \frac{\text{q}_{m}}{\varepsilon_{0}}$
Flux $\phi$through S'.
$\phi = \oint\limits_{s'}\overrightarrow{\text{E}}.\text{d}\overrightarrow{\text{S}} = \oint\limits_{s'}\text{EdS} = \text{E}.4\pi\text{r}^{2}$
$\Rightarrow\text{E}.4\pi\text{r}^{2} = \frac{\text{q}_{m}}{\varepsilon_{0}}\Rightarrow\text{E} = \frac{1}{4\pi\varepsilon_{0}}\frac{\text{q}_{m}}{\text{r}^{2}}$
View full question & answer→Question 403 Marks
A thin conducting spherical shell of radius R has charge Q spread uniformly over its surface. Using Gauss’s law, derive an expression for an electric field at a point outside the shell.Draw a graph of electric field E(r) with distance r from the centre of the shell for $0\underline{<}\text{r}\underline{<}\infty.$
Answer$\oint\overrightarrow{\text{E}}.\overrightarrow{\text{ds}} = \frac{\text{q}}{\varepsilon_\circ}$$\text{E}\times4\pi\text{r}^{2} =\frac{\text{Q}}{\varepsilon_\circ}$
$\text{E} = \frac{1}{4\pi\varepsilon_\circ}\frac{\text{Q}}{\text{r}^{3}}$
View full question & answer→Question 413 Marks
What is electric flux? Write its $S.I.$ units.Using Gauss’s theorem, deduce an expression for the electric field at a point due to a uniformly charged infinite plane sheet.
AnswerElectric Flux: Total number of electric lines of force crossing a certain area normally.
The surface integral of electric field over a closed surface.
Alternate Answer
$\phi = \oint\overrightarrow{\text{E}}.\text{d}\overrightarrow{\text{s}}$
- $S.I.$ Units: $Nm^2 /C$ Or $V-m$
-
Derivation: $\phi= \oint_{s}\overrightarrow{\text{E}}.\overrightarrow{\text{ds}}= \frac{\text{q}}{\varepsilon_\circ}$
$2\text{EA} =\frac{\sigma\text{A}}{\varepsilon_\circ}$
$\therefore \text{E} =\frac{\sigma}{2\varepsilon_{\circ}}$ View full question & answer→Question 423 Marks
Estimate the number of electrons in $100g$ of water. How much is the total negative charge on these electrons?
AnswerMolecular mass of water $= 18g$ Number of molecules in $18g$ of $\ce{H_2O} =$
Avogadro's number $= 6.023 \times 10^{23}$
Number of electrons in $1$ molecule of $\ce{H_2O} = (2 \times 1) + 8 = 10$
Number of electrons in $6.023 \times 10^{23}$ molecules of $\ce{H_2O} = 6.023 \times 10^{24}$
That is, number of electrons in $18g$ of $\ce{H_2O} = 6.023 \times 10^24$
So, number of electrons in $100g$ of $\ce{H_2O} =\frac{6.023\times10^{24}}{18}\times100$
$=3.34\times10^{25}$
$\therefore$ Total charge $=3.34\times10^{25}\times1.6\times10^{-19}$
$=5.34\times10^6\text{C}$
View full question & answer→Question 433 Marks
A charge $Q$ is uniformly distributed over a rod of length l. Consider a hypothetical cube of edge l with the centre of the cube at one end of the rod. Find the minimum possible flux of the electric field through the entire surface of the cube.
AnswerGiven: Total charge on the rod $= Q$
The length of the rod $=$ edge of the hypothetical cube $= l$
Portion of the rod lying inside the cube, $\text{x}=\frac{\text{l}}2{}$
Linear charge density for the rod $=\frac{\text{Q}}{\text{l}}$
Using Gauss's theorem, flux through the hypothetical cube,$\phi=\Big(\frac{\text{Q}_{\text{in}}}{\epsilon_0}\Big),$
where $Q_{in}=$ charge enclosed inside the cube
Here, charge per unit length of the rod $=\frac{\text{Q}}{\text{l}}$
Charge enclosed, $\text{Q}_{\text{in}}=\frac{\text{Q}}{\text{l}}\times\frac{\text{l}}{2}=\frac{\text{Q}}{2}$ Therefore,$\phi=\frac{\frac{\text{Q}}{2}}{\epsilon_0}=\frac{\text{Q}}{2\epsilon_0}$
View full question & answer→Question 443 Marks
Four point charges $q_A = 2 \mu C, q_B = –5 \mu C, q_C = 2 \mu C,$ and $q_D = –5 \mu C$ are located at the corners of a square $\text{ABCD}$ of side $10 \ cm$. What is the force on a charge of $1 \mu C$ placed at the centre of the square?
AnswerThe given figure shows a square of side $10 \ cm$ with four charges placed at its corners. $O$ is the centre of the square.
Where,
$($Sides$) AB = BC = CD = AD \ cm$
$($Diagonals$) AC = BD =10\sqrt{2} \ cm$
$AO = OC = DO = OB \ cm \ 5\sqrt{2} \ cm$
A charge of amount $1\mu C$ is placed at point $O$.
Force of repulsion between charges placed at corner $A$ and centre $O$ is equal in magnitude but opposite in direction relative to the force of repulsion between the charges placed at corner $C$ and centre $O$.
Hence, they will cancel each other.
Similarly, force of attraction between charges placed at corner $B$ and centre $O$ is equal in magnitude but opposite in direction relative to the force of attraction between the charges placed at corner $D$ and centre $O$.
Hence, they will also cancel each other.
Therefore, net force caused by the four charges placed at the corner of the square on $1\mu C$ charge at centre $O$ is zero. View full question & answer→Question 453 Marks
Two isolated metal spheres A and B have radii R and 2R respectively, and same charge q. Find which of the two spheres have greater energy density just outside the surface of the spheres.
AnswerEnergy density,$\text{U}=\frac{1}{2}\epsilon_0\text{E}^2$
But, $\text{E}=\frac{\sigma}{\epsilon_0}=\frac{\text{Q}}{\text{A}\epsilon_0}$$\therefore\ \text{U}=\frac{1}{2}\frac{\epsilon_0\text{Q}^2}{\text{A}^2\epsilon_0}\ \Rightarrow\ \text{U}=\frac{\text{Q}^2}{2\text{A}^2}$
$\Rightarrow\ \text{U}\propto\frac{1}{\text{A}^2}\ \Rightarrow\ \text{U}_\text{A}>\text{U}_\text{B}$
View full question & answer→Question 463 Marks
Two equal charges are placed at a separation of 1.0m. What should be the magnitude of the charges so that the force between them equals the weight of a 50kg person?
AnswerLet the magnitude of each charge be Separation between them, r = 1m Force between them, F = 50 × 9.8 = 490N By Coulomb's Law force,$\text{F}=\frac{1}{4\pi\in_0}\frac{\text{q}_1\text{q}_2}{\text{r}^2}$
$\Rightarrow490=9\times10^9\times\frac{\text{q}^2}{1^2}$
$\Rightarrow\text{q}^2=54.4\times10^{-9}$
$\Rightarrow\text{q}=\sqrt{54.4\times10^{-9}}$
$=23.323\times10^{-5}\text{C}$
$\text{q}=2.3\times10^{-4}\text{C}$
View full question & answer→Question 473 Marks
Find the speed of the electron in the ground state of a hydrogen atom. The description of ground state is given in the previous problem.
AnswerFe from previous problem No. $18=8.2\times10^{-8}\text{N},\ \text{Ve}=?$ Now, $\text{M}_\text{e}=9.12\times10^{31}\text{kg},\ \text{r}=0.53\times10^{-10}\text{m}$ Now, $\text{Fe}=\frac{\text{M}_\text{e}\text{v}^2}{\text{r}}$$\Rightarrow\text{v}^2=\frac{\text{Fe}\times\text{r}}{\text{m}_\text{e}}$
$=\frac{8.2\times10^{-8}\times0.53\times10^{-10}}{9.1\times10^{-31}}$
$=0.4775\times10^{13}$
$=4.775\times10^{12}\text{m}^2/\text{s}^2$
$\text{v}=2.18\times10^6\text{m/s}$
View full question & answer→Question 483 Marks
A charge of 1.0C is placed at the top of your college building and another equal charge at the top of your house. Take the separation between the two charges to be 2.0km. Find the force exerted by the charges on each other. How many times of your weight is this force?
AnswerGiven:$\text{q}_1=\text{q}_2=\text{q}=1.0\text{C}$
Distance between the charges, $\text{r}=2\text{km}=2\times10^3\text{m}$ By Coulomb's Law, electrostatic force,$\text{F}=\frac{1}{4\pi\in_0}\frac{\text{q}_1\text{q}_2}{\text{r}^2}$
$\text{F}=9\times10^9\times\frac{1\times1}{(2\times10^3)^2}$
$=2.25\times10^3\text{N}$
Let my mass, m, be 50kg. Weight of my body, W = mg$\Rightarrow\text{W}=50\times10\text{N}=500\text{N}$
Now,$\frac{\text{Weight of my body}}{\text{Force between the charges}}=\frac{500}{2.25\times10^3}$
$=\frac{1}{4.5}$
So, the force between the charges is 4.5 times the weight of my body.
View full question & answer→Question 493 Marks
The kinetic energy of a charged particle decreases by 10J as it moves from a point at potential 100V to a point at potential 200V. Find the charge on the particle.
AnswerK.C. decreases by 10J. Potential = 100v to 200v. So, change in K.E = amount of work done$\Rightarrow10\text{J}=(200-100)\text{v}\times\text{q}_0$
$\Rightarrow100\text{q}_0=10\text{v}$
$\Rightarrow\text{q}_0=\frac{10}{100}$
$=0.1\text{C}$
View full question & answer→Question 503 Marks
Does the force on a charge due to another charge depend on the charges present nearby?
AnswerCoulomb's Law states that the force between two charged particle is given by, $\text{F}=\frac{\text{q}_1\text{q}_2}{4\pi\in_0\text{r}^2}$
Where $,q_1$ and $q_2$ are the charges on the charged particles .
$r =$ separation between the charged particles. $\in_0 =$ parmittivity of free space.
According to the Law of Superposition, the electrostatic force between two charged particles are unaffected due to the presence of other charges.
View full question & answer→Question 513 Marks
A charged particle having a charge of $-2.0 \times 10^{-6}C$ is placed close to a nonconducting plate having a surface charge density $4.0 \times 10^{-5}Cm^{-2}.$ Find the force of attraction between the particle and the plate.
AnswerThe electric field due to a conducting thin sheet,$\text{E}=\frac{\sigma}{2\epsilon_0}$
The magnitude of attractive force between the particle and the plate,$\text{F}=\text{qE}$
$\text{F}=\frac{\text{q}\times\sigma}{2\epsilon_0}$
$\text{F}=\frac{\big(2.0\times10^{-6}\big)\times\big(4.0\times10^{-6}\big)}{2\times\big(8.85\times10^{-12}\big)}$
$\text{F}=0.45\text{N}$
View full question & answer→Question 523 Marks
A spherical volume contains a uniformly distributed charge of density $2.0 \times 10^{-4}Cm^{-3}.$ Find the electric field at a point inside the volume at a distance $4.0\ cm$ from the centre.
AnswerGiven : Volume charge density, $\rho=2\times10^{-4}\text{ C/m}^3$ Let us assume a concentric spherical surface inside the given sphere with radius $= 4\ cm = 4 \times 10^{-2}m$ The charge enclosed in the spherical surface assumed can be found by multiplying the volume charge density with the volume of the sphere. Thus,$\text{q}=\rho\times\frac{4}3{}\pi\text{r}^3$
$\Rightarrow\text{q}=\big(2\times10^{-4}\big)\times\frac{4}{3}\pi\text{r}^3$
The net flux through the spherical surface,$\phi=\frac{\text{q}}{\epsilon_0}$
The surface area of the spherical surface of radius $r\ cm:\text{A}=4\pi\text{r}^2$
Electric field,$\text{E}=\frac{\text{q}}{\epsilon_0\times\text{A}}$
$\text{E}=\frac{2\times10^{-4}\times4\pi\text{r}^3}{\epsilon_0\times3\times4\pi\text{r}^2}$
$\text{E}=\frac{2\times10^{-4}\times\text{r}}{3\times\epsilon_0\times}$
The electric field at the point inside the volume at a distance $4.0\ cm$ from the centre,$\text{E}=\frac{(2\times10^{-4})\times(4\times10^{-2})}{3\times(8.85\times10^{-12})}\text{N/C}$
$\text{E}=3.0\times10^{5}\text{ N/C}$
View full question & answer→Question 533 Marks
Two particles, carrying charges -q and +q and having equal masses m each, are fixed at the ends of a light rod of length a to form a dipole. The rod is clamped at an end and is placed in a uniform electric field E with the axis of the dipole along the electric field. The rod is slightly tilted and then released. Neglecting gravity find the time period of small oscillations.
AnswerConsider the rod to be a simple pendulum. For simple pendulum, 
$\text{T}=2\pi\sqrt{\frac{\ell}{\text{g}}}$ $\big(\ell=$ length, q = acceleration$\big)$
Now, force experienced by the charges F = Eq Now, acceleration $=\frac{\text{F}}{\text{m}}$$=\frac{\text{Eq}}{\text{m}}$
Hence length = a so, Time period $=2\pi\sqrt{\frac{\text{a}}{\Big(\frac{\text{Eq}}{\text{m}}\Big)}}$$=2\pi\sqrt{\frac{\text{ma}}{\text{Eq}}}$ View full question & answer→Question 543 Marks
Define electric field intensity. Write its $SI$ unit. Write the magnitude and direction of electric field intensity due to an electric dipole of length $2a$ at the midpoint of the line joining the two charges.
AnswerElectric Field Intensity: The electric field intensity at any point in an electric field is defined as the electric force per unit positive test charge placed at that point
i.e.,$\vec{\text{E}}=\lim\limits_{\text{q}_0\rightarrow0}\frac{\vec{\text{F}}}{\text{q}_0}$
The test charge $q0$ has to be vanishingly small so that it does not affect the electric field of the main charge.
The $SI$ unit of electric field intensity is newton/coulomb.
Electric Field Strength at mid $-$ point of dipole: The electric field strength at mid $-$ point $C$ due to charge $+q$ is $-q$ along the same direction.
$E = E_1 + E_2=\frac{1}{4\pi\epsilon_0}\frac{\text{q}}{\text{a}^2}+\frac{1}{4\pi\epsilon_0}\frac{\text{q}}{\text{a}^2}=\frac{1}{4\pi\epsilon_0}\frac{2\text{q}}{\text{a}^2}$
Its direction is from $+q$ to $-q$. View full question & answer→Question 553 Marks
A circular ring of radius r made of a nonconducting material is placed with its axis parallel to a uniform electric field. The ring is rotated about a diameter through $180^\circ$ . Does the flux of electric field change? If yes, does it decrease or increase?
AnswerIt is given that the circular ring, made of a non-conducting material, of radius r is placed with its axis parallel to a uniform electric field.This means that both the electric field and the area vector are parallel to each other $($area vector is always perpendicular to the surface area$)$. Thus, the flux through the ring is given by $\vec{\text{E}}.\vec{\text{S}}={\text{ES}}\cos0=\text{E}(\pi\text{r}^2).$
Now, when the ring is rotated about its diameter through $180^\circ$ the angle between the area vector and the electric field becomes $180^\circ $. Thus, the flux becomes $-\text{E}(\pi\text{r}^2).$
View full question & answer→Question 563 Marks
It is said that any charge given to a conductor comes to its surface. Should all the protons come to the surface? Should all the electrons come to the surface? Should all the free electrons come to the surface?
AnswerProtons never take part in any electrical phenomena because they are inside the nuclei and are not able to interact easily. These are the free electrons that are responsible for all electrical phenomena. So, if a conductor is given a negative charge, the free electrons come to the surface of the conductor. If the conductor is given a positive charge, electrons move away from the surface and leave a positive charge on the surface of the conductor.
View full question & answer→Question 573 Marks
The number of electrons in an insulator is of the same order as the number of electrons in a conductor. What is then the basic difference between a conductor and an insulator?
AnswerThe outer electrons of an atom or molecule in a conductor are only weakly bound to it and are free to move throughout the body of the material.
On the other hand, in insulators, the electrons are tightly bound to their respective atoms and cannot leave their parent atoms and move through a long distance.
View full question & answer→Question 583 Marks
When a charged comb is brought near a small piece of paper, it attracts the piece. Does the paper become charged when the comb is brought near it?
AnswerWhen a charged comb is brought near a small piece of paper, it attracts the piece due to induction. There's a distribution of charges on the paper. When a charged comb is brought near the pieces of paper then an opposite charge is induced on the near end of the pieces of paper so the charged comb attracts the opposite charge on the near end of paper and similar on the farther end. The net charge on the paper remains zero.
View full question & answer→Question 593 Marks
A spherical shell made of plastic, contains a charge Q distributed uniformly over its surface. What is the electric field inside the shell? If the shell is hammered to deshape it without altering the charge, will the field inside be changed? What happens if the shell is made of a metal?
AnswerAs the shell is made of plastic, it is non-conducting. But as the charge is distributed uniformly over the surface of the shell, the sum of all the electric field vectors at the centre due to this kind of distribution is zero. But when the plastic shell is deformed, the distribution of charge on it becomes non-uniform. In other words, the sum of all the electric field vectors is non-zero now or the electric field exists at the centre now.
In case of a deformed conductor, the field inside is always zero.
View full question & answer→Question 603 Marks
If a charge is placed at rest in an electric field, will its path be along a line of force? Discuss the situation when the lines of force are straight and when they are curved.
AnswerIf a charge is placed at rest in an electric field, its path will be tangential to the lines of force. When the electric field lines are straight lines then the tangent to them will coincide with the electric field lines so the charge will move along them only. When the lines of force are curved, the charge moves along the tangent to them.
View full question & answer→Question 613 Marks
Draw the electric field lines due to a uniformly charged thin spherical shell when charge on the shell is (a) positive and (b) negative.
AnswerThe electric field lines are shown in the figure. For a positively charged shell, the field lines are directed in radially outward direction and for negatively charged shell, these are directed in radially inward direction.
View full question & answer→Question 623 Marks
Two insulating small spheres are rubbed against each other and placed 1cm apart. If they attract each other with a force of 0.1N, how many electrons were transferred from one sphere to the other dunng rubbing?
Answer$\text{F}=0.1\text{N}$$\text{r}=1\text{cm}=10^{-2}$ (As they rubbed with each other. So the charge on each sphere are equal)
So, $\text{F}=\frac{\text{kq}_1\text{q}_2}{\text{r}^2}$
$\Rightarrow0.1=\frac{\text{kq}^2}{(10^{-2})^2}$
$\Rightarrow\text{q}^2=\frac{0.1\times10^{-4}}{9\times10^9}$
$\Rightarrow\text{q}^2=\frac{1}{9}\times10^{-14}$
$\Rightarrow\text{q}=\frac{1}{3}\times10^{-7}$
$1.6\times10^{-19}\text{c}$ Carries by 1 electron
1 c carried by $\frac{1}{1.6\times10^{-19}}$
$0.33\times10^{-7}$ c carries by
$\frac{1}{1.6\times10^{-19}} \times0.33\times10^{-7}$
$=0.208\times10^{12}$
$=2.08\times10^{11}$
View full question & answer→Question 633 Marks
Two identical particles, each having a charge of $2.0 \times 10^{-4}C$ and mass of $10g,$ are kept at a separation of $10\ cm$ and then released. What would be the speeds of the particles when the separation becomes large?
Answer
$\text{m}=10\text{g}$
$\text{F}=\frac{\text{KQ}}{\text{r}}$
$=\frac{9\times10^9\times2\times10^{-4}}{10\times10^{-2}}$
$\text{F}=1.8\times10^{-7}$
$\text{F}=\text{m}\times\text{a}$
$\Rightarrow\text{a}=\frac{1.8\times10^{-7}}{10\times10^{-3}}$
$=1.8\times10^{-3}\text{m/s}^2$
$\text{V}^2-\text{u}^2=2\text{as}$
$\Rightarrow\text{V}^2=\text{u}^2+2\text{as}$
$\text{V}=\sqrt{0+2\times1.8\times10^{-3}\times10\times10^{-2}}$
$=\sqrt{3.6\times10^{-4}}$
$=0.6\times10^{-2}$
$=6\times10^{-3}\text{,m/s}.$ View full question & answer→Question 643 Marks
Which among the curves shown in figure cannot possibly represent electrostatic field lines?
Answer
- Field lines are wrongly drawn because electric field lines must be normal to the surface of the conductor at each point.
- Field lines are wrongly drawn because field lines cannot start from a negative charge.
- Field lines are correctly drawn, because they are originating from a positive charge.
- Field lines are wrongly drawn as the field lines cannot intersect.
- Field lines are wrongly drawn because they cannot form closed loops.
View full question & answer→Question 653 Marks
Three equal charges, $2.0 \times 10^{-6}C$ each, are held fixed at the three corners of an equilateral triangle of side $5\ cm.$ Find the Coulomb force experienced by one of the charges due to the rest two.
Answer
Three charges are held at three corners of a equilateral trangle.
Let the charges be $A, B$ and $C.$
It is of length $5\ cm$ or $0.05m$
Force exerted by $B$ on $A = F_1$
Force exerted by $C$ on $A = F_2$
So, force exerted on $A =$ resultant $F_1 = F_2$
$\Rightarrow\text{F}=\frac{\text{kq}_2}{\text{r}^2}$
$=\frac{9\times10^9\times2\times2\times2\times10^{-12}}{5\times5\times10^{-4}}$
$=\frac{36}{25}\times10$
$=14.4$
Now, force on $A = 2 \times F \cos 30^\circ$ since it is equilateral $\triangle.$
$\Rightarrow$ Force on $\text{A}=2\times1.44\times\sqrt{\frac{3}{2}}$
$=24.94\text{N}.$ View full question & answer→Question 663 Marks
Two particles have equal masses of $5.0g$ each and opposite charges of $ +4.0 \times 10^{-5}C$ and $-4.0 \times 10^{-5}C$. They are released from rest with a separation of $1.0m$ between them. Find the speeds of the particles when the separation is reduced to $50\ cm$.
Answer
$\text{q}_1=\text{q}_2=4\times10^{-5}$
$\text{s}=1\text{m},\ \text{m}=5\text{g}$
$=0.005\text{kg}$
$\text{F}=\text{K}\frac{\text{q}^2}{\text{r}^2}$
$=\frac{9\times10^9\times\big(4\times10^{-5}\big)^2}{1^2}$
$=14.4\text{N}$
Acceleration ‘$a’ =\frac{\text{F}}{\text{m}}$
$=\frac{14.4}{0.005}$
$=2880\text{m/s}^2$
Now, $\text{u}=0,$
$\text{s}=50\text{cm}=0.5\text{m}$
$\text{a}=2880\text{m/s}^2,\ \text{V}=?$
$\Rightarrow\text{V}=\sqrt{2880}$
$=53.66\text{m/s}$
$\approx54\text{m/s}$ for each particle. View full question & answer→Question 673 Marks
What is the nature of electrostatic force between two point electric charges $q_1$ and $q_2$ if:
- $q_1 + q_2 > 0?$
- $q_1 + q_2 < 0?$
Answer
- If both $q_1$ and $q_2$ are positive, the electrostatic force between these will be repulsive.
However, if one of these charges is positive and is greater than the other negative charge, the electrostatic force between them will be attractive.
Thus, the nature of force between them can be repulsive or attractive.
- If both $q_1$ and $q_2$ are $-ve,$ the force between these will be repulsive.
However, if one of them is $-ve$ and it is greater in magnitude than the second $+ve$ charge, the force between them will be attractive.
Thus, the nature of force between them can be repulsive or attractive. View full question & answer→Question 683 Marks
A water particle of mass $10.0\ mg$ and having a charge of $1.50 \times 10^{-6}C$ stays suspended in a room. What is the magnitude of electric field in the room? What is its direction?
Answer
$\text{m}=10,\ \text{mg}=10\times10^{-3}\text{g}\times10^{-3}\text{kg},$
$\text{q}=1.5\times10^{-6}\text{C}$
But $\text{qE}=\text{mg}$
$\Rightarrow(1.5\times10^{-6})\text{E}=10\times10^{-6}\times10$
$\Rightarrow\text{E}=\frac{10\times10^{-4}\times10}{1.5\times10^{-6}}$
$=\frac{100}{1.5}=66.6\text{N/C}$
$=\frac{100\times10^3}{1.5}=\frac{10^{5+1}}{15}$
$=6.6\times10^{3}$ View full question & answer→Question 693 Marks
Three charges are arranged on the vertices of an equilateral triangle as shown in figure. Find the dipole moment of the combination.
Answer
Let -q & -q are placed at A & C
Where 2q on B
So length of A = d
So the dipole moment = (q × d) = P
So, Resultant dipole moment
$\text{P}=\Big[(\text{qd})^2+(\text{qd})^2+2\text{qd}\times\text{qd}\cos60^\circ\Big]^{\frac{1}{2}}$
$=\big[3\text{q}^2\text{d}^2\big]^{\frac{1}{2}}$
$=\sqrt{3}\text{qd}$
$=\sqrt{3}\text{p}$ View full question & answer→Question 703 Marks
A 10cm long rod carries a charge of $+50\mu\text{C}$ distributed uniformly along its length. Find the magnitude of the electric field at a point 10cm from both the ends of the rod.
Answer
$\text{G}=50\mu\text{C}=50\times10^{-6}\text{C}$
We have, $\text{E}=\frac{2\text{KQ}}{\text{r}}$ for a charged cylinder
$\Rightarrow\text{E}=\frac{2\times9\times10^9\times50\times10^{-6}}{5\sqrt{3}}$
$=\frac{9\times10^{-5}}{5\sqrt{3}}$
$=1.03\times10^{-5}$ View full question & answer→Question 713 Marks
Given a uniformly charged plane/sheet of surface charge density, $\sigma=2\times10^{17}\text{C/m}^2.$Image
- Find the electric field intensity at a point A, 5mm away from the sheet on the left side.
- Given a straight line with three points X, Y and Z placed 50cm away from the charged sheet on the right side. At which of these points, the field due to the sheet remain the same as that of point A and why?
Answer
- At A, $\text{E}=\frac{\sigma}{2\epsilon_0}=\frac{2\times10^{17}\text{Cm}^{-2}}{2\times8.854\times10^{-12}\text{C}^2\text{N}^{-1}\text{m}^{-2}}$
$\text{E}=1.1\times1028 \text{N/C}$
Directed away from the sheet.
- Point Y, Because at 50cm, the charge sheet acts as a finite sheet and thus the magnitude remains same towards the middle region of the planar sheet.
View full question & answer→Question 723 Marks
A particle having a charge of $2.0 \times 10^{-4}C$ is placed directly below and at a separation of $10\ cm$ from the bob of a simple pendulum at rest. The mass of the bob is $100g.$ What charge should the bob be given so that the string becomes loose?
Answer
Mass of the bob $= 100g = 0.1\ kg$
So Tension in the string $= 0.1 \times 9.8 = 0.98N.$
For the Tension to be $0,$ the charge below should repel the first bob.
$\Rightarrow\text{F}=\frac{\text{kq}_1\text{q}_2}{\text{r}^2}$ $\big[\text{T}-\text{mg}+\text{F}=0\ \Rightarrow\text{T}=\text{mg}-\text{f},\ \text{T}=\text{mg}\big]$
$\Rightarrow0.98=\frac{9\times10^9\times2\times10^{-4}\times\text{q}^2}{(0.01)^2}$
$\Rightarrow\text{q}_2=\frac{0.98\times1\times10^{-2}}{9\times2\times10^5}$
$=0.054\times10^{-9}\text{N}$ View full question & answer→Question 733 Marks
Three identical metal plates with large surface areas are kept parallel to each other as shown in figure. The leftmost plate is given a charge Q, the rightmost a charge -2Q and the middle one remains neutral. Find the charge appearing on the outer surface of the rightmost plate.
Answer
Consider the Gaussian surface as shown in the figure.
Let the charge on the outer surface of the left-most plate be q. Thus, the charges on the plates are distributed as shown in the diagram.
The net field at point P due to all the induced charges must be zero, as it is lying inside the metal surface.
Let the surface area of the plates be A.
Electric field at point P due to the charges on plate X:
Due to charge (+Q - q) is $\frac{\text{Q}-\text{q}}{2\text{A}\epsilon_0}$ in the right direction
Due to charge (+q) is $\frac{\text{q}}{2\text{A}\epsilon_0}$ in the right direction
Electric field at point P due to charges on plate Y:
Due to charge (-q) is $\frac{\text{q}}{2\text{A}\epsilon_0}$ in the right direction
Due to charge (+q) is $\frac{\text{q}}{2\text{A}\epsilon_0}$ in the left direction
Electric field at point P due to charges on plate Z:
Due to charge (-q) is $\frac{\text{q}}{2\text{A}\epsilon_0}$ in the right direction
Due to charge (-2Q + q) is $\frac{\text{2Q}-\text{q}}{2\text{A}\epsilon_0}$ in the right direction
The net electric field at point P:
$\frac{\text{Q}-\text{q}}{2\text{a}\epsilon_0}+\frac{\text{q}}{2\text{A}\epsilon_0}-\frac{\text{q}}{2\text{A}\epsilon_0}-\frac{\text{q}}{2\text{A}\epsilon_0}+\frac{\text{q}}{2\text{A}\epsilon_0}+\frac{\text{2Q}-\text{q}}{2\text{A}\epsilon_0}=0$
$\frac{\text{Q}-\text{q}}{2\text{A}\epsilon_0}+\frac{\text{2Q}-\text{q}}{2\text{A}\epsilon_0}=0$
$\text{Q}-\text{q}+\text{2Q}-\text{q}=0$
$\text{3Q}-2\text{q}=0$
$\text{q}=\frac{\text{3Q}}{2}$
Thus, the charge on the outer plate of the right-most plate
$-2\text{Q}+\text{q}=-2\text{Q}+\frac{3\text{Q}}{2}=-\frac{\text{Q}}{2}$ View full question & answer→Question 743 Marks
Two particles $A$ and $B$ having charges of $+2.00 \times 10^{-6}C$ and of $-4.00 \times 10^{-6}C$ respectively are held fixed at a separation of $20.0\ cm.$ Locate the point$(s)$ on the line $AB$ where
- The electric field is zero.
- The electric potential is zero.
Answer$\text{q}_2=2\times10^{-6}\text{C},\ \text{q}_1^2=-4\times10^{-6}\text{C},$$\text{r}=20\ cm=0.2\text{m}$
$(E_1 =$ electric field due to $q_1, E_2 =$ electric field due to $q_2)$
$\Rightarrow\frac{(\text{r}-\text{x})^2}{\text{x}^2}=\frac{-\text{q}_2}{\text{q}_1}$
$\Rightarrow\frac{(\text{r}-1)^2}{\text{x}}=\frac{-\text{q}_2}{\text{q}_1}$
$=\frac{4\times10^{-6}}{2\times10^{-6}}=\frac{1}{2}$
$\Rightarrow\Big(\frac{\text{r}}{\text{x}}-1\Big)=\frac{1}{\sqrt{2}}$
$=\frac{1}{1.414}$
$\Rightarrow\frac{\text{r}}{\text{x}}=1.414+1$
$=2.414$
$\Rightarrow\text{x}=\frac{\text{r}}{2.414}$
$=\frac{20}{2.414}$
$=8.285\ cm$
View full question & answer→Question 753 Marks
A charge $q$ is placed at the centre of the line joining two equal charges $Q.$
Show that the system of three charges will be in equilibrium if $\text{q}=-\frac{\text{Q}}{4}.$
Answer
Charge $q$ is in equilibrium since charges $A$ and $B$ exert equal and opposite forces on it.
For equilibrium of charge $Q$ at $B;$
$F_{BC} + F_{AB} = 0$
$\Rightarrow\ \frac{1}{4\pi\epsilon_0}\frac{\text{qQ}}{(\text{l}/2)^2}+\frac{1}{4\pi\epsilon_0}\frac{\text{Q.Q}}{\text{l}^2}=0$
$\Rightarrow\ \frac{1}{4\pi\epsilon_0}\frac{\text{Q}}{\text{l}^2}(4\text{q}+\text{Q})=0$
$\Rightarrow\ \text{}q=-\frac{\text{Q}}{4}$ View full question & answer→Question 763 Marks
Assume that each atom in a copper wire contributes one free electron. Estimate the number of free electrons in a copper wire having a mass of $6.4g\ ($take the atomic weight of copper to be $64g\ mol^{-1}.$
Answer$64$ grams of copper have $1$ mole
$6.4$ grams of copper have $0.1$ mole
$1$ mole $=$ No atoms
$0.1$ mole $= (no \times 0.1)$ atoms
$= 6 \times 10^{23}\times 0.1$ atoms $= 6 \times 10^{22}$ atoms
$1$ atom contributes $1$ electron
$6 \times 10^{22}$ atoms contributes $6 \times 10^{22}$ electrons.
View full question & answer→Question 773 Marks
Consider two hollow concentric spheres, $S_1$ and $S_2,$ enclosing charges $2Q$ and $4Q$ respectively as shown in the figure.
- Find out the ratio of the electric flux through them.
- How will the electric flux through the sphere $S_1$ change if a medium of dielectric constant $'\epsilon_\text{r}\ '$ is introduced in the space inside $S_1$ in place of air? Deduce the necessary expression.
AnswerUsing Gauss's Theorem $\oint\vec{\text{E}}.\vec{\text{ds}}=\frac{\text{q(T)}}{\epsilon_0}$
Electric flux through sphere $S_1,$ $\Phi_1=\frac{2(\text{Q})}{\epsilon_0}$
Electric flux through sphere $S_2, \Phi=\frac{(2\text{Q}+4\text{Q})}{\epsilon_0}=\frac{6\text{Q}}{\epsilon_0}$
$\text{Ratio}=\frac{\Phi_1}{\Phi}=\frac{\frac{2\text{Q}}{\epsilon_0}}{\frac{6\text{Q}}{\epsilon_0}}=\frac{1}{3}$
If a medium of dielectric constant $\text{K}(=\epsilon_\text{r})$ is filled in the sphere $S_1,$ electric flux through sphere,
$\Phi'_1=\frac{2\text{Q}}{\epsilon_\text{r}\epsilon_0}=\frac{2\text{Q}}{\text{K}\epsilon_0}.$
View full question & answer→Question 783 Marks
A nonconducting sheet of large surface area and thickness d contains uniform charge distribution of density $\rho.$ Find the electric field at a point P inside the plate, at a distance x from the central plane. Draw a qualitative graph of E against x for 0 < x < d.
AnswerGiven: Thickness of the sheet = d Let the surface area of the sheet be s. Volume of the sheet = sd Volume charge density of the sheet, $\rho=\frac{\text{Q}}{\text{sd}}$ Charge on the sheet = Q
Consider an imaginary plane at a distance x from the central plane of surface area s. Charge enclosed by this sheet, $\text{q}=\rho\text{sx}$ For this Guassian surface, using Gauss's Law,we get:$\oint\text{E.ds}=\frac{\text{q}}{\epsilon_0}$
$\text{E}.\text{s}=\frac{\rho\text{sx}}{\epsilon_0}$
$\text{E}=\frac{\rho\text{x}}{\epsilon_0}$
The electric field outside the sheet will be constant and will be:$\text{E}=\frac{\rho\text{d}}{\epsilon_0}$
View full question & answer→Question 793 Marks
Suppose an attractive nuclear force acts between two protons which may be written as $\text{F}=\text{Ce}^{-\text{kr}}/\text{r}^2.$
- Write down the dimensional formulae and appropriate $SI$ units of $C$ and $k.$
- Suppose that $k = 1$ fermi$^{-1}$ and that the repulsive electric force between the protons is just balanced by the attractive nuclear force when the separation is $5$ fermi. Find the value of $C.$
AnswerExpression of electrical force $\text{F}=\text{C}\times\text{e}^{\frac{-\text{kr}}{\text{r}^2}}$
Since $e^{-kr}$ is a pure number.
So, dimensional formulae of $\text{F}=\frac{\text{dimensional formulae of C}}{\text{dimensional formulae of r}^2}$
Or, $\big[\text{MLT}^{-2}\big]\big[\text{L}^2\big]=$ dimensional formulae of $\text{C}=\big[\text{ML}^3\text{T}^{-2}\big]$
Unit of $C =$ unit of force $\times$ unit of $r^2 =$ Newton $\times$ $m^2 =$ Newton$-m^2$
Since $-kr$ is a number hence dimensional formulae of $\text{k}=\frac{1}{\text{dim entional formulae of r}}=\big[\text{L}^{-1}\big]$ Unit of $k = m^{-1}$
View full question & answer→Question 803 Marks
A hollow charged conductor has a tiny hole cut into its surface. Show that the electric field in the hole is $(\sigma /2ε_0) \hat{\text{n}}$ , where $\hat{\text{n}}$ is the unit vector in the outward normal direction, and $\sigma$ is the surface charge density near the hole.
AnswerLet us take a charged conductor with the hole filled up, as shown by shaded portion in the figure. 
We find with the application of Gaussian theorem that field inside is zero and just outside is $\frac{\sigma}{\epsilon_0}\hat{\text{n}}.$ This field can be viewed as the superposition of the field $E_2$ due to the filled up hole plus the field $E_1$ due to the rest of the charged conductor. The two fields $(E_1$ and $E_2)$ must be equal and opposite as the field vanishes inside the conductor. Thus, $E_1 - E_2 = 0$
Now, the field outside the conductor is given by$\text{E}_1+\text{E}_2=\frac{\sigma}{\epsilon_0}$
$\therefore \ 2\ \text{E}_1=\frac{\sigma}{\epsilon_0}$
$\Rightarrow\ \text{E}_1=\frac{\sigma}{2\epsilon_0}$
Therefore, field in the hole $($due to the rest of the conductor$)$ is
given as:$\text{E}_1=\frac{\sigma}{2\epsilon_0}\hat{\text{n}}$ ($\hat{\text{n}} \rightarrow$ unit vector in the outward normal direction$)$ View full question & answer→Question 813 Marks
Consider a sphere of radius R with charge density distributed as$\rho\text{(r)}=\text{kr for r}\leq\text{R}$
$=0\text{ for f}>\text{R}.$
Find the electric field at all points r.
AnswerThe expression of charge density distribution in the sphere suggests that the electric field is radial. Let us consider a sphere S of radius R and two hypothetic spheres of radius r < R and r > R. Let us first consider for point r < R, electric field intensity will be given by, 
$\oint\overrightarrow{\text{E}}.\text{d}\overrightarrow{\text{S}}=\frac{1}{\epsilon_0}\int\rho\text{dV}$
Here $\text{dV}=4\pi\text{r}^2\text{dr}$$\Rightarrow\ \oint\overrightarrow{\text{E}}.\text{d}\overrightarrow{\text{S}}=\frac{1}{\epsilon_0}4\pi\text{K}\int\text{r}^3\text{dr}\ \ (\because\ \rho(\text{r})=\text{Kr})$
$\Rightarrow\ (\text{E})4\pi\text{r}^2=\frac{4\pi\text{K}}{\epsilon_0}\frac{\text{r}^4}{4}$
We get, $\text{E}=\frac{1}{4\epsilon_0}\text{Kr}^2$ As charge density is positive, it means the direction of E is radially outwards. Now consider points r > R, electric field intensity will be given by$\oint\overrightarrow{\text{E}}.\text{d}\overrightarrow{\text{S}}=\frac{1}{\epsilon_0}\int\rho\text{dV}$
$\Rightarrow\ \text{E}(4\pi\text{r}^2)=\frac{4\pi\text{K}}{\epsilon_0}\oint\text{r}^3\text{dr}=\frac{4\pi\text{K}}{\epsilon_0}\frac{\text{R}^4}{4}$
Which given, $\text{E}=\frac{\text{K}}{4\epsilon_0}\frac{\text{r}^4}{\text{r}^2}$ Here also the charge density is again positive. So, the direction of E is radially outward. View full question & answer→Question 823 Marks
Consider the situation shown in figure. What are the signs of $q_1$ and $q_2\ ?$ If the lines are drawn in proportion to the charge, what is the ratio $\frac{\text{q}_1}{\text{q}_2}?$
AnswerThe electric lines of force are entering charge $q_1;$
So, it is is negative.
On the other hand, the lines of force are originating from charge $q_2;$ so, it positive.
If the lines are drawn in proprotion to the charges, then$\frac{\text{q}_1}{\text{q}_2}=\frac{6}{18}$
$\Rightarrow\frac{\text{q}_1}{\text{q}_2}=\frac{1}{3}$
$6$ lines are entering $q_1$ and $18$ are coming our of $q_2.$
View full question & answer→Question 833 Marks
The electric field in a region is given by $\overrightarrow{\text{E}}=\frac{3}{5}\text{E}_0\overrightarrow{\text{i}}+\frac{4}{5}\text{E}_0\overrightarrow{\text{j}}$ with $\text{E}_0=2.0\times10^3\text{NC}^{-1}.$ Find the flux of this field through a rectangular surface of area $0.2m^2$ parallel to the $y-z$ plane.
AnswerGiven: Electric field strength, $\overrightarrow{\text{E}}=\frac{3}{5}\text{E}_0\hat{\text{i}}+\frac{4}{5}\text{E}_0\hat{\text{j}}$
where $\text{E}_0=2.0\times10^3\text{N/C}$ The plane of the rectangular surface is parallel to the $y-z$ plane.
The normal to the plane of the rectangular surface is along the $x$ axis.
Only $\frac{3}{5}\text{E}_0\hat{\text{i}}$ passes perpendicular to the plane;
so, only this component of the field will contribute to flux.
On the other hand, $\frac{4}{5}\text{E}_0\hat{\text{i}}$ moves parallel to the surface.
Surface area of the rectangular surface, $a = 0.2m^2$ Flux,$\phi=\overrightarrow{\text{E}}.\overrightarrow{\text{a}}=\text{E}\times\text{a}$
$\phi=\Big(\frac{3}5{}\times2\times10^3\Big)\times(2\times10^{-1})\text{Nm}^2/\text{C}$
$\phi=0.24\times10^3\text{Nm}^2/\text{C}$
$\phi=240\text{Nm}^2/\text{C}$
View full question & answer→Question 843 Marks
Two identically charged particles are fastened to the two ends of a spring of spring constant $100Nm^{-1}$ and natural length $10\ cm.$ The system rests on a smooth horizontal table. If the charge on each particle is $2.0 \times 10^{-8}C,$ find the extension in the length of the spring. Assume that the extension is small as compared to the natural length. Justify this assumption after you solve the problem.
Answer$\text{K}=100\text{N/m},\ \ell=10\text{cm}=10^{-1}\text{m} \text{q}=2.0\times10^{-8}\text{c},\ \text{Find}\ \ell=?$
Force between them $\text{F}=\frac{\text{kq}_1\text{q}_2}{\text{r}^2}$
$=\frac{9\times10^92\times10^{-8}\times2\times10^{-8}}{10^{-2}}$
$=36\times10^{-5}\text{N}$
So, $\text{F}=-\text{kx}$ or $\text{x}=\frac{\text{F}}{-\text{K}}$
$=\frac{36\times10^{-5}}{100}$
$=36\times10^{-7}\text{ cm}$
$=3.6\times10^{-6}\text{m}$ View full question & answer→Question 853 Marks
By the Gauss's law, derive the expression of the electric field at any point due to a similar charged straight wire of infinite length. Draw necessary diagram.
Question 863 Marks
Due to the electric dipole, drive the formula of the electric field at any point located on its axial line. Draw necessary diagram.
