3 Marks Question - Physics STD 12 Science Questions - Vidyadip

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Question 13 Marks

Find equivalent capacitance between A and B in the combination given below. Each capacitor is of 2 μF capacitance.

If a dc source of 7 V is connected across AB, how much charge is drawn from the source and what is the energy stored in the network?

Answer

Capacitors $C_2$, $C_3$ $\text{and}$ $C_4 $$\ \ \text{are in parallel}$
$\therefore C_{234}=C_2+C_3+C_4$
$\therefore C_{234}=6\mu F$
Capacitors $C_1$,$C_{234}$ $\text{and}$ $C_5$ are in series
$\frac{1}{C_{eq}}=\frac{1}{C_1}+\frac{1}{C_{234}}+\frac{1}{C_5}=\frac{1}{2}+\frac{1}{6}+\frac{1}{2}$
$=\frac{7}{6}\mu F$
$C_{equivalent}=\frac{6}{7}\mu F$
Charge drawn from the source
$Q=C_{eq}V,$
$=\frac{6}{7}\times7\mu C=6\mu C$
Energy stored $U=\frac{Q^2}{2C}$
$=\frac{6\times6\times10^{-12}\times7}{2\times6\times10^{-6}}\text{J}$
$=21\mu \text{J}$.

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Question 23 Marks

A spherical conducting shell of inner radius $r_1$ and outer radius $r_2$ has a charge Q.

A charge q is placed at the centre of the shell. What is the surface charge density on the inner and outer surfaces of the shell?
Is the electric field inside a cavity (with no charge) zero, even if the shell is not spherical, but has any irregular shape? Explain

Answer

Charge placed at the centre of a shell is +q. Hence, a charge of magnitude -q will be induced to the inner surface of the shell. Therefore, total charge on the inner surface of the shell is -q.

Surface charge density at the inner surface of the shell is given by the relation,

$\sigma_1=\frac{\text{Total charge}}{\text{Inner surface area}}=\frac{-\text{q}}{4\pi\text{r}^2_1} \dots\dots(1)$

A charge of +q is induced on the outer surface of the shell. A charqe of magnitude Q is placed on the outer surface of the shell. Therefore, total charge on the outer surface of the shell is Q + q. Surface charge density at the outer surface of the shell,

$\sigma_2=\frac{\text{Total charge}}{\text{Outer surface area}}=\frac{-\text{Q+q}}{4\pi\text{r}^2_2} \dots\dots(2)$

Yes

The electric field intensity inside a cavity is zero, even if the shell is not spherical and has any irregular shape. Take a closed loop such that a part of it is inside the cavity along a field line while the rest is inside the conductor. Net work done by the field in carrying a test charge over a closed loop is zero because the field inside the conductor is zero. Hence, electric field is zero, whatever is the shape.

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Question 33 Marks

A parallel plate capacitor with air between the plates has a capacitance of 8 pF $(1pF = 10^{–12}F)$. What will be the capacitance if the distance between the plates is reduced by half, and the space between them is filled with a substance of dielectric constant 6?

Answer

Capacitance between the parallel plates of the capacitor, C = 8 pF
Initially, distance between the parallel plates was d and it was filled with air. Dielectric constant of air, k = 1
Capacitance, C, is given by the formula,
$\text{C}=\frac{\text{k}\in_0\text{A}}{\text{d}}$
$=\frac{\in_0\text{A}}{\text{d}} \dots\dots(1)$
Where,
A = Area of each plate
$∈_0=$ Permittivity of free space
If distance between the plates is reduced to half, then new distance, $\text{d}^{-\text{TM}}=\frac{\text{d}}{2}$ Dielectric constant of the substance filled in between the plates, k'= 6 Hence, capacitance of the capacitor becomes
$\text{C}'=\frac{\text{k}'\in_0\text{A}}{\text{d}}=\frac{6\in_0\text{A}}{\frac{\text{d}}{2}} \dots\dots(2)$
Taking ratios of equations (i) and (ii), we obtain
$C' = 2 × 6C$
$= 12\ C$
$= 12 × 8 = 96pF$
Therefore, the capacitance between the plates is 96 pF.

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Question 43 Marks

A spherical conductor of radius 12 cm has a charge of 1.6 × 10–7C distributed uniformly on its surface. What is the electric field:

Inside the sphere
Just outside the sphere
At a point 18 cm from the centre of the sphere?

Answer

Radius of the spherical conductor, $r = 12\ cm = 0.12\ m$
Charge is uniformly distributed over the conductor, $q = 1.6 \times 10^{-7}\ C$

Electric field inside a spherical conductor is zero. This is because if there is field inside the conductor, then charges will move to neutralize it.
Electric field E just outside the conductor is given by the relation,

$\text{E}=\frac{\text{q}}{4\pi\in_0\text{r}^2}$
Where,
$∈_0 =$ Permittivity of free space
$\frac{1}{4\pi\in_0}=9\times10^9\text{Nm}^2\text{C}^{-1}$
$\therefore\text{E}=\frac{1.6\times10^{-7}\times9\times10^{-9}}{(0.12)^2}$
$= 10^5 NC^{-1}$
Therefore, the electric field just outside the sphere is $10^5 NC^{-1}.$

Electric field at a point 18 m from the centre of the sphere = E1

Distance of the point from the centre, d = 18 cm = 0.18 m
$\text{E}_1=\frac{\text{q}}{4\pi\in_0\text{d}^2}$
$=\frac{9\times10^9\times1.6\times10^{-7}}{18\times10^{-2}}$
$= 4.4 \times 10^4 N/C$
Therefore, the electric field at a point 18 cm from the centre of the sphere is $4.4 \times 10^4 N/C.$

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Question 53 Marks

A spherical capacitor consists of two concentric spherical conductors, held in position by suitable insulating supports (2.36). Show

that the capacitance of a spherical capacitor is given by $\text{C}=\frac{{4}\pi\in_{0}\text{r}_1\text{r}_2}{\text{r}_1-\text{r}_2}$ where $r_1$ and $r_2$ are the radii of outer and inner spheres, respectively.

Answer

Radius of the outer shell $= r_1$
Radius of the inner shell $= r_2$
The inner surface of the outer shell has charge +Q.
$\text{V}=\frac{\text{Q}}{{4}\pi\in_0\text{r}_2}-\frac{Q}{{4}\pi\in_0\text{r}_1}$
The outer surface of the inner shell has induced charge −Q. Potential difference between the two shells is given by,
Where,
$∈_0 =$ Permittivity of free space
$\text{V}=\frac{\text{Q}}{{4}\pi\in_0}\bigg[\frac{1}{\text{r}_2}-\frac{1}{\text{r}_1}\bigg]$
$\text{V}=\frac{\text{Q}(\text{r}_1-\text{r}_2)}{{4}\pi\in_0\text{r}_1\text{r}_2}$
Capacitance of the given system is given by,
$\text{C}\frac{\text{charge}(\text{Q})}{\text{potential difference}(\text{V})}$
$=\frac{{4}\pi\in_0\text{r}_1\text{r}_2}{\text{r}_1-\text{r}_2}$
Hence, proved.

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Question 63 Marks

If one of the two electrons of a $H_2$ molecule is removed, we get a hydrogen molecular ion $\text{H}^+_2.$ In the ground state of an $\text{H}^+_2,$ the two protons are separated by roughly 1.5 Å, and the electron is roughly 1 Å from each proton. Determine the potential energy of the system. Specify your choice of the zero of potential energy.

Answer

The system of two protons and one electron is represented in the given figure.

Charge on proton $1, q_1 = 1.6 x 10^{-19} C$
Charge on proton $2, q_2= 1.6 x 10^{-19} C$
Charge on electron, $q_3 = -1.6 x 10-^{19} C$
Distance between protons $1$ and $2, d_1 = 1.5 x 10^{-10} m$
Distance between proton $1$ and electron,$ d_2 = 1 x 10^{-10} m$
Distance between proton $2$ and electron, $d_3 = 1 x 10^{-10} m$
The potential energy at infinity is zero.
Potential energy of the system, $\text{V}=\frac{\text{q}_1\text{q}_2}{4\pi\in_0\text{d}_1}+\frac{\text{q}_2\text{q}_3}{4\pi\in_0\text{d}_1}$

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Question 73 Marks

A regular hexagon of side 10 cm has a charge 5 µC at each of its vertices. Calculate the potential at the centre of the hexagon.

Answer

The given figure shows six equal amount of charges, q, at the vertices of a regular hexagon.

Where,
Charge, $q = 5 µC = 5 \times 10^{-6} C$
Side of the hexagon,$ l = AB = BC = CD = DE = EF = FA = 10 cm$
Distance of each vertex from centre $O, d = 10 cm$
Electric potential at point $O,$
$\text{V}=\frac{6\times q}{4\pi\in_0\text{d}}$
Where,
$∈_0 =$ Permittivity of free space
$\frac{1}{4\pi\in_0}=9\times10^9\text{NC}^{-2}\text{m}^{-2}$
$\therefore\text{V}=\frac{6\times9\times10^9\times5\times10^{-6}}{0.1}$
$= 2.7 \times 10^6V$
Therefore, the potential at the centre of the hexagon is $2.7 \times 10^6 V.$

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Question 83 Marks

In a hydrogen atom, the electron and proton are bound at a distance of about 0.53 $\mathring{\text{A}}$:

Estimate the potential energy of the system in eV, taking the zero of the potential energy at infinite separation of the electron from proton.
What is the minimum work required to free the electron, given that its kinetic energy in the orbit is half the magnitude of potential energy obtained in (a)?
What are the answers to (a) and (b) above if the zero of potential energy is taken at 1.06 Å separation?

Answer

The distance between electron-proton of a hydrogen atom, d = 0.53 $\mathring{\text{A}}$
Charge on an electron, $\mathrm{q}_1=-1.6 \times 10^{-19} \mathrm{C}$
Charge on a proton, $\mathrm{q}_2=+1.6 \times 10^{-19} \mathrm{C}$

Potential at infinity is zero.

Potential energy of the system, p-e = Potential energy at infinity - Potential energy at distance, d

$=0-\frac{\text{q}_1\text{q}_2}{4\pi\in_0\text{d}}$

Where,

$\in_0$ is the permittivity of free space

$\frac{1}{4\pi\in_0}=9\times10^9\text{Nm}^2\text{C}^{-2}$

$\therefore\text{Potential energy}=0-\frac{9\times10^9\times\big(1.6\times10^{-19}\big)^2}{0.53\times10^{-10}}=-43.7\times10^{-19}\text{J}$

Since $1.6x 10^{-19} J = 1\ eV,$

$\therefore\text{Potential energy}=-43.7\times10^{-19}=\frac{-43.7\times10^{-19}}{1.6\times10^{-19}}=-27.2\ \text{eV}$

Therefore, the potential energy of the system is -27.2 ev.

Kinetic energy is half of the rnagnitude of potential energy.

Kinetic energy $=\frac{1}{2}\times(-27.2)=13.6\ \text{eV}$

Total energy $= 13.6 - 27.2 = 13.6\ eV$

Therefore, the rninimum work required to free the electron is 13.6 eV.

When zero of potential energy is taken, $d_1= 1.06\ A$

$\therefore$ Potential energy of the system = Potential energy at $d_1 -$ Potential energy at d

$=\frac{\text{q}_1\text{q}_2}{4\pi\in_0\text{d}_1}-27.2\ \text{eV}$

$=\frac{9\times10^9\times(1.6\times10^{-19})^2}{1.06\times10^{-10}}-27.2\ \text{eV}$

$= 21.73 \times 10^{-19}\ J - 27.2\ eV$

$= 13.58 eV - 27.2\ eV$

$= -13.6\ eV$

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Question 93 Marks

In a parallel plate capacitor with air between the plates, each plate has an area of $6 \times 10^{–3} m^2$ and the distance between the plates is 3 mm. Calculate the capacitance of the capacitor. If this capacitor is connected to a 100 V supply, what is the charge on each plate of the capacitor?

Answer

Area of each plate of the parallel plate capacitor, $A = 6 \times 10^{-3} m^2$
Distance between the plates, $d = 3 mm = 3 \times 10^{-3} m$
Supply voltage, $V = 100\ V$
Capacitance C of a parallel plate capacitor is given by,
$\text{C}=\frac{\in_0\text{A}}{\text{d}}$
Where,
$\in_0$ = Permittivity of free space
$= 8.854 \times 10^{-12} N^{-1}m^{-2}C^{-2}$
$\therefore\text{C}=\frac{8.854\times10^{-12}\times6\times10^{-3}}{3\times10^{-3}}$
$= 17.71 \times 10^{-12}\ F$
$= 17.71pF$
Potential V is related with the charge q and capacitance C as
$\text{V}=\frac{\text{q}}{\text{C}}$
$\therefore\text{q}=\text{VC}$
$= 100 \times 17.71 \times 10^{-12}$
$= 1.771 \times 10^{-9}\ C$
Therefore, capacitance of the capacitor is 17.71 pF and charge on each plate is $1.771 \times 10^{-9}.$

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Question 103 Marks

What is the area of the plates of a 2F parallel plate capacitor, given that the separation between the plates is 0.5 cm? [You will realise from your answer why ordinary capacitors are in the range of μF or less. However, electrolytic capacitors do have a much larger capacitance (0.1F) because of very minute separation between the conductors.]

Answer

Capacitance of a parallel capacitor, V = 2FDistance between the two plates, $d = 0.5cm = 0.5 \times 10^{-2}m$ Capacitance of a parallel plate capacitor is given by the relation,
Where,
$\in_{0}= \text{Permittivity of free space} = {8.85}\times{10}^{-12}\text{C}^{2}\text{N}^{-1}\text{m}^{-2}$
$\therefore\ \text{A} = \frac{2\times0.5\times10^{-2}}{8.85\times10^{-12}}$
$={1130}\text{km}^{2}$
Hence, the area of the plates is too large. To avoid this situation, the capacitance is taken in the range of μF.

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Question 113 Marks

Three capacitors of capacitances 2 pF, 3 pF and 4 pF are connected in parallel.

What is the total capacitance of the combination?
Determine the charge on each capacitor if the combination is connected to a 100 V supply.

Answer

Capacitances of the given capacitors are

$C_1= 2\ pF$
$C_2 = 3\ pF$
$C_3 = 4\ pF$
For the parallel combination of the capacitors, equivalent capacitor r C is given by the algebraic sum,
C' = 2 + 3 + 4 = 9 pF
Therefore, total capacitance of the combination is 9 pF.

Supply voltage, V = 100 V

The voltage through all the three capacitors is same = V = 100 V
Charge on a capacitor of capacitance C and potential difference V is given by the relation,
q = vc ......(1)
For C = 2 pF,
Charge $= VC = 100 \times 2 = 200 pc = 2 \times 10^{-10}\ C$
For C = 3 pF,
Charge $= VC = 100 \times 3 = 300 pc = 3 \times 10^{-10}\ C$
For C = 4 pF,
Charge $= VC = 100 \times 4 = 400 pc = 4 \times 10^{-10}\ C$

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Question 123 Marks

Two charged conducting spheres of radii a and b are connected to each other by a wire. What is the ratio of electric fields at the surfaces of the two spheres? Use the result obtained to explain why charge density on the sharp and pointed ends of a conductor is higher than on its flatter portions.

Answer

Let a be the radius of a sphere $A, Q_A$ be the charge on the sphere,
and $C_A$ be the capacitance of the sphere.
Let b be the radius of a sphere $B, Q_B$ be the charge on the sphere,
and $C_B$ be the capacitance of the sphere. Since the two spheres are connected with a wire,
their potential (V) will become equal. Let $E_A$ be the electric field of sphere A and $E_B$ be the electric field of sphere B.
Therefore, their ratio,$\frac{\text{E}_{\text{A}}}{\text{E}_{\text{B}}}=\frac{\text{Q}_{\text{A}}}{{4}\pi\in_{0} \text{a}^2}\times\frac{\text{b}^2{4}\pi\in_{0}}{\text{Q}_{\text{B}}}$
$\frac{\text{E}_{\text{A}}}{\text{E}_{\text{B}}}=\frac{\text{Q}_\text{A}}{\text{Q}_\text{B}}\times \frac{\text{b}^{2}}{\text{a}^{2}} \ \ ...(1)$
However, $\frac{\text{Q}_\text{A}}{\text{Q}_\text{B}}=\frac{\text{C}_\text{A}\text{V}}{\text{C}_\text{B}\text{V}}$ And, $\frac{\text{C}_\text{A}}{\text{C}_\text{B}}=\frac{\text{a}}{\text{b}}$ $\therefore \ \frac{\text{Q}_\text{A}}{\text{Q}_\text{B}}=\frac{\text{a}}{\text{b}}\ \ ...(2)
$Putting the value of (2) in (1), we obtain
$\therefore \ \frac{\text{E}_\text{A}}{\text{E}_\text{B}}=\frac{\text{a}}{\text{b}}\frac{\text{b}^{2}}{\text{a}^{2}}=\frac{\text{b}}{\text{a}}$
Therefore, the ratio of electric fields at the surface is $\frac{\text{b}}{\text{a}}$.

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Question 133 Marks

Explain what would happen if in the capacitor given in Exercise 2.8, a 3 mm thick mica sheet (of dielectric constant = 6) were inserted between the plates,

While the voltage supply remained connected.
After the supply was disconnected.

Answer

Dielectric constant of the mica sheet, k = 6

Initial capacitance, $C = 1.771 \times 10^{-11}$

New capacitance, $C^r = kC = 6 \times 1.771 \times 10^{-11} = 106\ pF$

Supply voltage, $V = 100\ V$

New charge $q^1 = C' = 6 \times 1.7717 \times 10^{-9} = 1.06 \times 10^{-8}\ Cs$

Potential across the plates remains 100 V.

Dielectric constant, $k = 6$

Initial capacitance $C' = kC = 6 \times 1.771 \times 10^{-11} = 106\ pF$

If supply voltage is removed, then there will be no effect on the amount of charge in the plates.

Charge $= 1.771 \times 10^{-9}\ C$

Potential across the plates is given by,

$\therefore\text{V}'=\frac{\text{q}}{\text{C}'}$

$=\frac{1.771\times10^{-9}}{106\times10^{-12}}$

$= 16.7\ V$

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Question 143 Marks

A long charged cylinder of linear charged density λ is surrounded by a hollow co-axial conducting cylinder. What is the electric field in the space between the two cylinders?

Answer

Charge density of the long charged cylinder of length L and radius r is $\lambda.$
Another cylinder of same length surrounds the pervlous cylinder. The radius of this cylinder is R.
Let E be the electric field produced in the space between the two cylinders.
Electric flux through the Gaussian surface is given by Gauss's theorem as,
$\phi=\text{E}(2\pi\text{d})\text{L}$
Where, d = Distance of a point from the common axis of the cylinders
Let q be the total charge on the cylinder.
It can be written as
$\therefore\phi=\text{E}(2\pi\text{dL})=\frac{\text{q}}{\in_0}$
Where,
q = Charge on the inner sphere of the outer cylinder
$\in_0$ = Permittivity of free space
$\text{E}(2\pi\text{dL})=\frac{\lambda\text{L}}{\in_0}$
$\text{E}=\frac{\lambda}{2\pi\in_0\text{d}}$
Therefore, the electric field In the space between the two cylinders is $\frac{\lambda}{2\pi\in_0\text{d}}.$

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Question 153 Marks

A cylindrical capacitor has two co-axial cylinders of length 15cm and radii 1.5cm and 1.4cm. The outer cylinder isearthed and the inner cylinder is given a charge of 3.5µC. Determine the capacitance of the system and the potential of the inner cylinder. Neglect end effects (i.e., bending of field lines at the ends).

Answer

Length of a co-axial cylinder, I = 15cm= 0.15m Radius of outer cylinder, $r_1= 1.5cm = 0.015m$
Radius of inner cylinder, $r_2 = 1.4cm = 0.014m$
Charge on the inner cylinder, $q = 3.5µC = 3.5 x 10^{-6}\ c$
Capacitance of a co-axial cylinder of radii $r_1$ and $r_2$ is given by the relation,
$\text{C}=\frac{{2}\pi\in_{0}\text{I}}{\log\frac{\text{r}_1}{\text{r}_2}}$ Where,
$\in_0$ = Permittivity of free space $= 8.85 \times 10^{-12} N^{-1} m^{-2} C^2$
$\therefore\ \text{C}=\frac{{2}\pi\times{8.85}\times{10}^{-12}\times{0.15}}{{2.3026}\log_{10}\bigg(\frac{0.15}{0.14}\bigg)}$ $=\frac{{2}\pi\times{8.85}\times{10}^{-12}\times{0.15}}{{2.3026}\times{0.0299}}={1.2}\times{10}^{-10}\text{F}$
Potential difference of the inner cylinder is given by, $\text{V}=\frac{\text{q}}{\text{C}}$ $=\frac{3.5\times10^{-6}}{1.2\times10^{-10}}={2.92}\times{10}^{4}\text{V}$

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Question 163 Marks

Three circuits, each consisting of a switch ‘S’ and two capacitors, are initially charged, as shown in the figure. After the switch has been closed, in which circuit will the charge on the left-hand capacitor (i) increase, (ii) decrease and (iii) remain same? Give reasons.

Answer

$V_L = 3?$	$V_R = 3?$	(L: Left, R: Right)
$V_L = 6?$	$V_R = 3?$
$V_L = 2?$	$V_R= 3?$

Reasons:

No change – (potential same on both capacitors as $(V_L = V_R).$
Charge on left hand capacitor will decrease $(V_L > V_R).$
Charge on left hand capacitor will increase $(V_R > V_L).$

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Question 173 Marks

Define an equipotential surface. Draw equipotential surface:

In the case of a single point charge and,
In a constant electric field in Z-direction.

Why the equipotential surfaces about a single charge are not equidistant?

Can electric field exist tangential to an equipotential surface? Give reason.

Answer

Surface with a constant value of potential at all points on the surface.

$\text{V} \propto\frac{1}{\text{r}}/$

No If the field lines are tangential, work will be done in moving a charge on the surface which goes against the definition of equipotential surface.

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Question 183 Marks

Two identical parallel plate capacitors A and B are connected to a battery of V volts with the switch S closed. The switch is now opened and the free space between the plates of the capacitors is filled with a dielectric of dielectric constant K. Find the ratio of the total electrostatic energy stored in both capacitors before and after the introduction of the dielectric.

Answer

Energy stored $=\frac{1}{2}CV^2=(\frac{1}{2}\frac{Q^2}{C})$ Net capacitance with switch S closed = C + C = 2C$\therefore$ Energy stored $=\frac{1}{2}\times2C\times V^2=CV^2$
After the switch S is opened, capacitance of each capacitor = KC$\therefore$ Energy stored in capacitor A $=\frac{1}{2}KCV^2$
For capacitor B, Energy stored $=\frac{1}{2}\frac{Q^2}{KC}=\frac{1}{2}\frac{C^2V^2}{KC}=\frac{1}{2}\frac{CV^2}{K}$$\therefore$ Total Energy stored $=\frac{1}{2}{KCV^2}+\frac{1}{2}\frac{CV^2}{K}=\frac{1}{2}{CV^2}\bigg({K+\frac{1}{K}}\bigg)$
$=\frac{1}{2}{CV^2}\bigg({\frac{K^2+1}{K}}\bigg)$
$\therefore$ Required ratio $=\frac{\frac{1}{2}\text{KCV}^2}{\frac{1}{2}\text{CV}^2\Big(\frac{\text{K}^2+1}{\text{K}}\Big)}=\frac{\text{K}^2}{\text{K}^2+1}$

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Question 193 Marks

Deduce the expression for the electrostatic energy stored in a capacitor of capacitance 'C' and having charge 'Q'.
How will the (i) energy stored and (ii) the electric field inside the capacitor be affected when it is completely filled with a dielectric material of dielectric constant 'K'?

Answer

Potential difference between the plates of capacitor
V = q/C
Work done to add additional charge dq on the capacitor
dw = V x dq= (q/C) x dq
$\therefore$Total energy stored in the capacitor
$\text{U} = \int\text{dw} = \int_{0}^{Q}\frac{\text{q}}{\text{c}}\text{dq} = \frac{1}{2}\frac{\text{Q}^{2}}{\text{c}}$
When battery is disconnected:

Energy stored will be decreased or energy stored = 1/K times the initial energy.
Electric field would decrease or E’ = E/K.

Alternate Answer
When battery is connected:

Energy stored will increase or become K times the initial energy.
Electric field will not change.

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Question 203 Marks

A network of four capacitors each of 12 μF capacitance is connected to a 500 V supply as shown in the figure. Determine (a) equivalent capacitance of the network and (b) charge on each capacitor.

Answer

$\frac{1}{\text{C}} = \frac{1}{\text{C}_{1}} + \frac{1}{\text{C}_{2}} + \frac{1}{\text{C}_{3}}$

$\frac{1}{\text{C}} = \frac{1}{12} + \frac{1}{12} +\frac{1}{12} = \frac{1}{4}$

$\text{C} = 4\mu\text{F}\text{ or simply C}_\text{s} = \frac{\text{C}}{3} = \frac{12}{3}\mu\text{F} = 4 \mu\text{F}$

Equivalent capacitance

$\text{C}_{eq} = \text{C} + \text{C}_{4} = 4 +12 = 16 \mu\text{F}$

Calculation of charge on each capacitor:

Charge on capacitor $C_4$

$\text{Q}_{4} = \text{C}_{4}\text{V} = 12 \times500\mu\text{C} = 6000\mu\text{C} = 6 \times10^{-3}\text{C}$

Charge on capacitors $C_1, C_2$ and $C_3$

$\text{Q}_{123} =4 \mu\text{F}\times500\text{V} = 2 \times10^{-3}\text{C}. $

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Question 213 Marks

A parallel plate capacitor is charged by a battery. After some time the battery is disconnected and a dielectric slab of dielectric constant K is inserted between the plates. How would (i) the capacitance, (ii) the electric field between the plates and (iii) the energy stored in the capacitor, be affected? Justify your answer.

Answer

Original capacitance $\text{C}_{o} = \frac{\text{Q}}{\text{V}_{o}} = \varepsilon_{0}\frac{\it{\text{A}}}{\text{d}}$
When a dielectric is inserted:

Capacitance

$\bigg( = \text{K}\in_{o}\frac{\it{\text{A}}}{\text{d}}\bigg)\text{ increases}$

Electric Field.

$ = \bigg(\frac{\sigma - \sigma\text{P}}{\in_{o}}\bigg)\text{decreases}$

Energy stored

$\bigg(\text{W} = \frac{1}{2}\text{KC}_{o}\cdot\frac{\text{Q}^{2}}{\text{C}_{o}\text{K}^{2}} =\frac{1}{2}\frac{\text{Q}^{2}}{\text{C}_{o}}\cdot\frac{1}{\text{k}}\bigg)\text{decrease}.$

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Question 223 Marks

Explain the underlying principle of working of a parallel plate capacitor. If two similar plates, each of area A having surface charge densities $+\sigma$ and $ -\sigma$ are separated by a distance d in air, write expressions for.

the electric field at points between the two plates.
the potnetial difference between the plates.
the capacitance of the capacitor so formed.

Answer

Principle:

When an uncharged, grounded conducting plate is placed near a charged conducting plate a charge, of the opposite sign, gets induced on the
This reduces the potential of first plate, without any change in the charge present in it.

$\text{E} = \frac{\sigma}{\varepsilon_\circ}$

$\text{V}= \frac{\sigma}{\varepsilon_\circ}\text{d}$

$\text{C} =\frac{\varepsilon_\circ}{\text{d}}\text{A}$.

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Question 233 Marks

In the following arrangement of capacitors, the energy stored in the 6 μF capacitor is E. Find the value of the following:
Energy stored in 12 μF capacitor.
Energy stored in 3 μF capacitor.
Total energy drawn from the battery.

Answer

$\text{E}=\frac{1}{2}CV^2=\frac{6}{2}\times10^{-6}V^{2}=3\times10^{-6}V^2$

$\therefore V^2=\frac{E}{3\times10^{-6}}$
Energy stored in $12\mu f$ capacitor $=\frac{1}{2}CV^2$
$=\frac{1}{2}\times12\times10^{-6}\times\frac{E}{3\times10^{-6}}$
$= 2E$

Charge on $6\mu f$ capacitor, $Q_1=\sqrt{2EC}\big[\because E= \frac{1}{2}\frac{Q^2}{C}\big]$

$=2\sqrt{3E}\times10^{-3}C$
Charge on $12\mu f$ capacitor, $Q_2=\sqrt{2CE}$
$=\sqrt{2\times12\times10^{-6}\times2E}$
$=4\sqrt{3E}10^{-3}C$
Charge on $3\mu f$ apacitor, $Q = Q_1+ Q_2$
$=6\sqrt{3E}10^{-3}$
Energy stored in $3\mu f$ capacitor $=\frac{1}{2}\frac{Q^2}{C}=\frac{1}{2}\frac{36\times3E\times10^{-6}}{3\times10^{-6}}$
= 18E

Alternate Answer

capacitance of parallel combination = $18\mu f$

Charge on parallel combination, $Q = CV$
$=18\times10^{-6}V$
Charge on $3\mu f=\text{Q}=3\times10^{-6}V_1$
$=18\times10^{-6}V=3\times10^{-6}V_1$
$=V_1=6V$
$\therefore$ Energy stored in $3\mu f$ capacitor $=\frac{1}{2}CV_1^2$
$=\frac{1}{2}\times3\times10^{-6}\times\frac{E\times36}{3\times10^{-6}}$
= 18 E

Total energy drawn $= E + 2E + 18E = 21E$

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Question 243 Marks

Two identical capacitors of 12 pF each are connected in series across a battery of 50 V. How much electrostatic energy is stored in the combination? If these were connected in parallel across the same battery, how much energy will be stored in the combination now?
Also find the charge drawn from the battery in each case.

Answer

In series combination: $\frac{1}{C_S}=\Bigg(\frac{1}{12}+\frac{1}{12}\Bigg)(pF)^{-1}$
$\therefore C_s=6\times10^{-12}pF$
$U_S=\frac{1}{2}CV^2$
$U_S=\frac{1}{2}\times6\times 10^{-12}\times50\times50\text{ }\text{J}$
$\therefore U_s=75\times10^{-10}\text{ }\text{J}$
$q_s=C_ sV$
$6\times50$
$300\times10^{-12} C=3\times{10}^{-10}C$

In parallel combination: ${C}_{p}=(12+12)pF$
$\therefore{C}_{p}=24\times10^{-12}F$
$U_s=\frac{1}{2}\times24\times10^{-12}\times2500\text{ }\text{J}$
$=3\times10^{-8}\text{ }\text{J}$
$q_p=C{p} V$
$q_p=24\times10{}^{-12}\times50\text{ }\text{C}$
$q_p=1.2\times10^{-9}\text{ }\text{C}$

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Question 253 Marks

Derive the expression for the electric potential due to an electric dipole at a point on its axial line.
Depict the equipotential surfaces due to an electric dipole.

Answer

Potential due to charge at $ {A},{V}_{\text{A}}=\frac{1}{4{\pi}\in_0}\frac{-q}{(r+a)}$ Potential due to charge at $ {B},{V}_{B}=\frac{1}{4{\pi}\in_0}\frac{+q}{(r-a)}$

$ \therefore{\text{potential at point}}\text{ }P,V={V_B}+{V_A}$
$ \therefore{\text{ Net potential at}}\text{ }\text{P}=\frac{q}{4{\pi}\in_0}\bigg[\frac{-1}{(r+a)}+\frac{1}{(r-a)}\bigg]$
$V=\frac{q\times{2a}}{{4{\pi}\in_0}(r^2-a^2)}$

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Question 263 Marks

Two parallel plate capacitors X and Y have the same area of plates and same separation between them. X has air between the plates while Y contains a dielectric medium of $\varepsilon_{\text{r}} = 4.$

Calculate capacitance of each capacitor if equivalent capacitance of the combination is $4\mu\text{f}.$
Calculate the potential difference between the plates of X and Y.
Estimate the ratio of electrostatic energy stored in X and Y.

Answer

Let $\text{C}_{x} = \text{C}$

$\text{C}_{y} = 4\text{C } \text{as it has a dielectric medium of } \varepsilon_{r} = 4 .$

For series combination of two capacitors

$\frac{1}{\text{C}} =\frac{1}{\text{C}_{x}} + \frac{1}{\text{C}_{y}}$

$\Rightarrow\frac{1}{4\mu\text{F}} = \frac{1}{\text{C}} + \frac{1}{4\text{C}}$

$\frac{1}{4\mu\text{F}} = \frac{5}{4\text{C}}$

$\Rightarrow\text{C} = 5\mu\text{F}$

Hence $\text{C}_{X} = 5\mu\text{F}$

$\text{C}_{y} = 20 \mu\text{F}$

Total charge Q = CV

$ = 4 \mu\text{F}\times15 \text{V} = 60 \mu\text{C}$

$\text{V}_{x} = \frac{\text{Q}}{\text{C}_{x}} = \frac{60\mu\text{C}}{5\mu\text{F}} = 12 \text{V}$

$\text{V}_{y} = \frac{\text{Q}}{\text{C}_{y}} = \frac{60\mu\text{C}}{20\mu\text{F}} = 3 \text{V}$

$\frac{{\text{E}}_{\text{x}}}{\text{E}_{\text{y}}} = \frac{^\frac{\text{Q}^{2}}{2{C}_{\text{X}}}}{\frac{\text{Q}^{2}}{2\text{C}_{Y}}} = \frac{\text{C}_{\text{y}}}{\text{C}_{\text{x}}} = \frac{20}{5} = 4:1$

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Question 273 Marks

Find equivalent capacitance between A and B in the combination given below. Each capacitor is of 2 μF capacitance.

If a dc source of 7 V is connected across AB, how much charge is drawn from the source and what is the energy stored in the network?

Answer

Capacitors $C_2$, $C_3$ $\text{and}$ $C_4 $$\ \ \text{are in parallel}$$\therefore C_{234}=C_2+C_3+C_4$
$\therefore C_{234}=6\mu F$
Capacitors $C_1$,$C_{234}$ $\text{and}$ $C_5$ are in series$\frac{1}{C_{eq}}=\frac{1}{C_1}+\frac{1}{C_{234}}+\frac{1}{C_5}=\frac{1}{2}+\frac{1}{6}+\frac{1}{2}$
$=\frac{7}{6}\mu F$
$C_{equivalent}=\frac{6}{7}\mu F$
Charge drawn from the source$Q=C_{eq}V,$
$=\frac{6}{7}\times7\mu C=6\mu C$
Energy stored $U=\frac{Q^2}{2C}$$=\frac{6\times6\times10^{-12}\times7}{2\times6\times10^{-6}}\text{J}$
$=21\mu \text{J}$.

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Question 283 Marks

Two capacitors of unknown capacitances $C_1$ and $C_2$ are connected first in series and then in parallel across a battery of 100 V. If the energy stored in the two combinations is 0.045 J and 0.25 J respectively, determine the value of $C_1$ and $C_2$. Also calculate the charge on each capacitor in parallel combination.

Answer

Energy stored in a capacitor$\text{E} = \frac{1}{2}\text{CV}^{2}$

in series combination$0.045 =\frac{1}{2}\frac{\text{c}_{1}\text{c}_{2}}{\text{c}_{1}\text{c}_{2}}(100)^{2}$
$ = > \frac{\text{c}_{1}\text{c}_{2}}{\text{c}_{1} + \text{c}_{2}} = 0.09\times10^{-4}$.....(i)
In parallel combination$0.25 = \frac{1}{2}(\text{c}_{1} +\text{c}_{2})(100)^{2}$
$= > \text{c}_{1} + \text{c}_{2} =0.5\times10^{-4}$.....(ii)
On simplifying (i) & (ii)
$C_1C_2 = 0.045 \times 10^{–8} (C_1 – C_2)^{2}$
$= (C_1 + C_2)^2 – 4C_1C_2 = (0.5 \times 10^{–4})^2 – 4 \times 0.045 \times 10^{–8}$
$= 0.25 \times 10^{–8} – 0.180 \times 10^{–8} (C_1 – C_2)^2= 0.07 \times 10^{–8} (C_1 – C_2)^{2}$
$= 2.6 \times 10^{–5} = 0.26 \times 10^{–4} .....(iii)$ From (ii) and (iii)
we have $= > C_1 = 0.38 \times 10^{–4} F$ and $C_2 = 0.12 \times 10^{–4} F$
Charges on capacitor $C_1$ and $C_2$ in parallel combination. $Q_1= C_1V = (0.38 \times 10^{–4} \times 100)$
$= 0.38 \times 10^{–2}C Q_2= C_2V = (0.12 \times 10^{–4} \times 100) = 0.12 \times 10^{–2}C$
$\text{E} = \frac{1}{2}\text{CV}^{2}$
and $0.045 = \frac{1}{2}\bigg(\frac{\text{c}_{1}\text{c}_{2}}{\text{c}_{1} +\text{c}_{2}}\bigg)(100 )^{2}$
$0.25 = \frac{1}{2}(\text{c}_{1} +\text{c}_{2})(100)^{2}$
but is unable to calculate $C_1$ and $C_2, Q_1 = C_1V = C_1(100)$ and $Q_2 = C_2V = C_2(100).$

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Question 293 Marks

Depict the equipotential surfaces for a system of two identical positive point charges placed a distance ‘d’ apart.
Deduce the expression for the potential energy of a system of two point charges $q_1$ and $q_2$ brought from infinity to the points$\overrightarrow{\text{r}_{1}}$ and $\overrightarrow{\text{r}_{2}}$ respectively in the presence of external electric field $\overrightarrow{\text{E}}.$

Answer

Equipotential surfaces for a system of two identical positive charges:

Expression for the potential energy of a system of two point charges in external field: Work done in bringing the charge $q_1$ from infinity to $r_1$ Work done $= q_1 V(r_1)$ Work done in bringing the charge $q_1$ from infinity to $r_2.$ Work done against the external electric field $= q_2V (r_2)$ Work done = work done against the external electric field + Work done on $q_2$ against the field due to $q_1= \text{q}_{2}\text{V}(\text{r}_{2}) + \frac{\text{q}_{1}\text{q}_{2}}{4\pi\varepsilon_{0}\text{r}_{12}}$
Potential energy of the system
= the total work done in assembling the configuration$ = \text{q}_{1}\text{V}(\text{r}_{1}) + \text{q}_{2}\text{V}(\text{r}_{2}) + \frac{\text{q}_{1}\text{q}_{2}}{4\pi\varepsilon_{0}\text{r}_{12}}.$

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Question 303 Marks

A parallel plate capacitor is charged by a battery. After sometime the battery is disconnected and a dielectric slab with its thickness equal to the plate separation is inserted between the plates. How will (i) the capacitance of the capacitor, (ii) potential difference between the plates and (iii) the energy stored in the capacitor be affected?
Justify your answer in each case.

Answer

Capacltance $\text{C} = \frac{\text{K}\varepsilon_{0}\text{A}}{\text{d}},$ Hence capacitance Increases K times.
Potential difference $\text{V} = \frac{\text{V}_{o}}{\text{K}},$Hence potential difference decreases by a factor K.
Energy stored E $ =\frac{1}{2}\text{CV}^{2},$ As capacitance becomes K times & potential difference becomes 1/K times therefore energy stored becomes 1/K times.

Alternate Answer
Energy stored $= Q^2/2C$. As capacitance increases by a factor K, the energy stored will decrease by the same factor.

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Question 313 Marks

Three identical capacitors $C_1, C_2$ and $C_3$ of capacitance 6 µF each are connected to a 12 V battery as shown.

Find:

charge on each capacitor.
equivalent capacitance of the network.
energy stored in the network of capacitors.

Answer

Charge on capacitors $C_1$ and $C_{2.}$

$\text{Q}_{1} = 36\mu\text{C} $ ;

charge on capacitor $C_3$

$\text{Q}_{3} = 72\mu\text{C}$

Equivalent capacitance.

$\text{C} = \frac{\text{C}_{1}{\text{C}_{2}}}{\text{C}_{1}+\text{C}_{2}} + \text{C}_{3}$

$\frac{6\times6}{6+6}+6 = 9\mu\text{F}$

Energy stored in Network.

$\text{W} =\frac{1}{2}\text{CV}^{2}$

$ =\frac{1}{2}\times9\times10^{-6}\times\big(12\big)^{2}\text{J}$

$ =648\times10^{-6}\text{J} = 648\mu\text{J}$

$ = 6.48\times10^{-4}\text{J}$

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Question 323 Marks

A parallel plate capacitor, each with plate area A and separation d, is charged to a potential difference V. The battery used to charge it is then disconnected. A dielectric slab of thickness d and dielectric constant K is now placed between the plates. What change, if any, will take place in.

Charge on the plates.
Electric field intensity between the plates.
Capacitance of the capacitor.

Answer

No change.

As the battery is disconnected.

Decreases OR becomes $\frac{1}{\text{k}}$ times Due to polarisation of the dielectric.
Increases OR becomes k times.

As the electric field, and therefore, the p.d., between the plates decreases.

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Question 333 Marks

Deduce an expression for the electric potential due to an electric dipole at any point on its axis. Mention one contrasting feature of electric potential of a dipole at a point as compared to that due to a single charge.

Answer

Potential at P due to +q
$\text{V}_{+q} = \frac{1}{4\pi\varepsilon_{\circ}}$ $\frac{q}{\times-a}$
Potential at P due to -q
$\text{V}_{-q} = \frac{1}{4\pi\varepsilon_{\circ}}$ $\frac{-q}{\times+a}$
$\therefore$ Potential at P due to the dipole
$\text{V} = \frac{1}{4\pi\varepsilon_{\circ}}$ $\frac{q\big(2\text{a}\big)}{(\text{X}^{2}-\text{a}^{2})}$
for a <
$\therefore \text{v}\cong\frac{1}{4\pi\varepsilon_\circ}$ $\frac{q\big(2\text{a}\big)}{\text{X}^{2}}$
For a single charge V $\alpha\frac{1}{\text{X}^{2}}$

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Question 343 Marks

Four point charges Q, q, Q and q are placed at the corners of a square of side ‘a’ as shown in the figure.

Find the:

Resultant electric force on a charge Q.
Potential energy of this system.

Answer

Force on Q due to Q:

$\text{F}_1=\frac{\text{KQ}^2}{(\sqrt{2}\text{a})^2}=\frac{\text{KQ}^2}{2\text{a}^2}$ (direction is shown in figure)

Force on Q due to single 'q':

$\text{F}=\frac{\text{KQq}}{\text{a}^2}$ (direction is same as $F_1$)

Resultant Force on Q due to both 'q':

$\text{F}_2=\sqrt{2}\text{F}=\frac{\sqrt{2}\text{KQq}}{\text{a}^2}$

$\therefore$ Resulatant electric force on Q:

$\text{F}=\text{F}_1+\text{F}_2$

$\text{F}=\frac{\text{KQ}^2}{2\text{a}^2}+\frac{\sqrt{2}\text{KQq}}{\text{a}^2}$

$\text{F}=\frac{\text{KQ}}{\text{a}^2}\Big(\frac{\text{Q}}{2}+\sqrt{2}\text{q}\Big)$

$\text{F}=\frac{\text{KQ}}{\text{a}^2}\Big(\frac{\text{Q}+2\sqrt{2}\text{q}}{2}\Big)$

Directed outwards along the line joing charges Q and Q. (Show in figure)

Potential Energy of system:

$\text{U}=\frac{\text{KQq}}{\text{a}}+\frac{\text{KQq}}{\text{a}}+\frac{\text{KQ}^2}{\sqrt{2}\text{a}}+\frac{\text{KQq}}{\text{a}}+\frac{\text{Kq}^2}{\sqrt{2}\text{a}}+\frac{\text{KQq}}{\text{a}}$

$=\frac{4\text{KQq}}{\text{a}}+\frac{\text{KQ}^2}{\sqrt{2}\text{a}}+\frac{\text{Kq}^2}{\sqrt{2}\text{a}}$

$\text{P.E.}=\frac{\text{K}}{\text{a}}\Big[4\text{Qq}+\frac{\text{Q}^2}{\sqrt{2}}+\frac{\text{q}^2}{\sqrt{2}}\Big]$

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Question 353 Marks

Three point charges q, -4q and 2q are placed at the vertices of an equilateral triangle ABC of side ‘l’ as shown in the figure. Obtain the expression for the magnitude of the resultant electric force acting on the charge q.

Find out the amount of the work done to separate the charges at infinite distance.

Answer

Force on q due to -4q.

$\text{F}_1=\frac{\text{K(4q)(q)}}{\text{l}^2}$ (along A - B)

$\text{F}_1=\frac{\text{4Kq}^2}{\text{l}^2}$ (along A - B)

Force on q due to 2q.

$\text{F}_2=\frac{\text{K(2q)(q)}}{\text{l}^2}$ (along C - A)

$\text{F}_2=\frac{2\text{Kq}^2}{\text{l}^2}$ (along C - A)

Angle between $F_1$ of $F_2$ $(\theta=120^\circ):$

$\therefore$ Resultat Force:

$\text{F}=\sqrt{\text{F}^2_1+\text{F}^2_2+2\text{F}_1\text{F}_2\cos120^\circ}$

$\text{F}=\sqrt{(2\text{F}_2)^2+(\text{F}_2)^2+2(2\text{F}_2)\text{F}_2\Big(-\frac{1}{2}\Big)}$ $(\because\ \text{F}_1=2\text{F}_2)$

$=\sqrt{5\text{F}^2_2-2\text{F}^2_2}=\sqrt{3}\text{F}_2$

$\therefore\ \text{F}=\sqrt{3}\times\frac{2\text{Kq}^2}{\text{l}^2}$

Direction $\tan\alpha=\frac{\text{F}_2\sin\theta}{\text{F}_1+\text{F}_2\cos\theta}$

$=\frac{\text{F}_2\times\frac{\sqrt{3}}{2}}{2\text{F}_2+\text{F}_2\big(-\frac{1}{2}\big)}$

So, resultant forces is making angle.

$\alpha=\tan^{-1}\Big(\frac{1}{\sqrt{3}}\Big)$ with the line AB.

Workdone to gather the charges from infinity = potential energy of system (U)

$\text{U}=\frac{\text{Kq(2q)}}{\text{l}}+\frac{\text{Kq(-4q)}}{\text{l}}+\frac{\text{K(2q)(-4q)}}{\text{l}}$

$=\frac{\text{Kq}^2}{\text{l}}(2-4-8)=-10\frac{\text{Kq}^2}{\text{l}}$

$\therefore$ Workdone to separate the charges to infinity $=\frac{10\text{Kq}^2}{\text{l}}$

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Question 363 Marks

Draw the equipotential surfaces corresponding to a uniform electric field in the z-direction.
Derive an expression for the electric potential at any point along the axial line of an electric dipole.

Answer

For constant electric field vector E

For increasing electric field

Difference: For constant electric field, the equipotential surfaces are equidistant for the same potential difference between these surfaces; while for increasing electric field, the separation between these surfaces decreases, in the direction of the increasing field, for the same potential difference between them.

Suppose P is a point on the axial position of the dipole. Length of dipole = 2a

Suppose point P is at the distance 'r' from the center of the dipole.

We know that potential at a point is given as, $\text{V}=\frac{1}{4\pi\in_0}.\frac{\text{Q}}{\text{r}}$

So, the potential at P due to q is, $\text{V}_\text{q}=\frac{1}{4\pi\in_0}.\frac{\text{q}}{\text{a}+\text{r}}$

Potential at P due to -q is, $\text{V}_\text{-q}=\frac{1}{4\pi\in_0}.\frac{\text{-q}}{\text{a}-\text{r}}$

The total potential at P is,

$\text{V}=\text{V}_\text{q}+\text{V}_{-\text{q}}$

$=\frac{1}{4\pi\in_0}.\frac{\text{q}}{(\text{a}+\text{r})}+\frac{1}{4\pi\in_0}.\frac{-\text{q}}{(\text{r-a})}$

$=\frac{\text{q}}{4\pi\in_0}\Big[\frac{1}{(\text{a+r})}+\frac{1}{(\text{a-r})}\Big]$

$\text{V}=\frac{\text{q}}{4\pi\in_0}.\frac{2\text{a}}{(\text{a}^2-\text{r}^2)}$

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Question 373 Marks

A charge of $+2.0 \times 10^{-8}C$ is placed on the positive plate and a charge of $-1.0 \times 10^{-8}C$ on the negative plate of a parallel-plate capacitor of capacitance $1.2\times10^{-3}\mu\text{F}.$ Calculate the potential difference developed between the plates.

Answer

$\text{q}_1=+2.0\times10^{-8}\text{c}$$\text{q}_2=-1.0\times10^{-8}\text{c}$
$\text{C}=1.2\times10^{-3}\mu\text{F}=1.2\times10^{-9}\text{F}$
$\text{net q}=\frac{\text{q}_1-\text{q}_2}{2}=\frac{3.0\times10^{-8}}{2}$
$\text{V}=\frac{\text{q}}{\text{c}}=\frac{3\times10^{-8}}{2}\times\frac{1}{1.2\times10^{-9}}=12.5\text{V}$

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Question 383 Marks

A parallel-plate capacitor having plate area $25cm^2$ and separation 1.00mm is connected to a battery of 6.0V. Calculate the charge flown through the battery. How much work has been done by the battery during the process?

Answer

$\text{A}=25\text{cm}^2=2.5\times10^{-3}\text{cm}^2$
$\text{d}=1\text{mm}=0.01\text{m}$
$\text{V}=\text{6V},\ \text{Q}=1$
$\text{C}=\frac{\in_0\text{A}}{\text{d}}=\frac{8.854\times10^{-12}\times2.5\times10^{-3}}{0.01}$
$\text{Q}=\text{CV}=\frac{8.854\times10^{-12}\times2.5\times10^{-3}}{0.01}\times6$
$=1.32810\times10^{-10}\text{C}$
$\text{W}=\text{Q}\times\text{V}=1.32810\times10^{-10}\text{C}\times6$
$=8\times10^{-10}\text{J}.$

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Question 393 Marks

Find the total energy stored in the capacitors in the given network.

Answer

The equivalent capacitance of $C_1$ and $C_2$ in series,$\text{C}'=\frac{\text{C}_1\text{C}_2}{\text{C}_2+\text{C}_2}=\frac{2\times2}{2+2}=1\mu\text{F}$
C′ is in parallel with $C_3$, so equivalent capacitance of $C_1, C_2$ and $C_3$ is:$\text{C}''=1+1=2\mu\text{F}$
C′′ is in series with $C_4;$ their equivalent capacitance,$\text{C}'''=\frac{\text{C}_4\text{C}}{\text{C}_4+\text{C}''}=\frac{2\times2}{2+2}=1\mu\text{F}$
This is in parallel with $C_5$; So equivalent capacitance across AB is $\text{C}_{\text{AB}}=1+1=2\mu\text{F},$ Energy stored $\text{V}'=\frac{1}{2}\text{C}_{\text{AB}}\text{V}^2=\frac{1}{2}\times2\times10^{-6}\times(6)^2=36\times10^{-6}\text{J}$

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Question 403 Marks

Prove that a closed equipotential surface with no charge within itself must enclose an equipotential volume.

Answer

Let us assume that in a closed equipotential surface with no charge the potential is changing from position to position. Let the potential just inside the surface is different to that of the surface causing in a potential gradient $\Big(\frac{\text{dV}}{\text{dr}}\Big)$ It means $\text{E}\neq0$ electric field comes into existence, which is given by as $\text{E}=-\frac{\text{dV}}{\text{dr}}$ It means there will be field lines pointing inwards or outwards from the surface. These lines cannot be again on the surface, as the surface is equipotential. It is possible only when the other end of the field lines are originated from the charges inside. This contradicts the original assumption. Hence, the entire volume inside must be equipotential.

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Question 413 Marks

A charge of $20\mu\text{C}$ is placed on the positive plate of an isolated parallel-plate capacitor of capacitance $10\mu\text{F}.$ Calculate the potential difference developed between the plates.

Answer

$\therefore$ Given that
Capacitance $=10\mu\text{F}$
Charge $=20\mu\text{C}$
$\therefore$ The effective charge $=\frac{20-0}{2}=10\mu\text{F}$
$\therefore\text{C}=\frac{\text{q}}{\text{V}}$
$\Rightarrow\text{V}=\frac{\text{q}}{\text{C}}=\frac{10}{10}=1\text{V}$

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Question 423 Marks

Find the charges on the three capacitors connected to a battery as shown in figure. Take $\text{C}_1=2.0\mu\text{F},\ \text{C}_2=4.0\mu\text{F},\ \text{C}_3=6.0\mu\text{F}$ and V = 12 volts.

Answer

$\text{C}_1=2\mu\text{F},\ \text{C}_2=4\mu\text{F},\ \text{C}_3=6\mu\text{F}$
$\text{V}=12\text{V}$
$\text{cq}=\text{C}_1+\text{C}_2+\text{C}_3$
$=2+4+6=12\mu\text{F}$
$=12\times10^{-6}\text{F}$
$\text{q}_1=12\times2=24\mu\text{C}$
$\text{q}_2=12\times4=48\mu\text{C}$
$\text{q}_3=12\times6=72\mu\text{C}$

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Question 433 Marks

Concentric equipotential surfaces due to a charged body placed at the centre are shown. Identify the polarity of the charge and draw the electric field lines due to it.

Answer

For a single charge the potential is given by $\text{V}=\frac{1}{4\pi\varepsilon_0}\frac{\text{q}}{\text{r}}$
This shows that V is constant if r is constant. Greater the radius smaller will be the potential. In the given figure, potential is increasing. This shows that the polarity of charge is negative (-q). The direction of electric field will be radially inward. The field lines are directed from higher to lower potential.

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Question 443 Marks

Prove that, if an insulated, uncharged conductor is placed near a charged conductor and no other conductors are present, the uncharged body must be intermediate in potential between that of the charged body and that of infinity.

Answer

If we analyse the equation, $\text{E}=-\frac{\text{dV}}{\text{dr}}$, then we observe that electric potential decreases along the direction of electric field.
Now, consider any path from the charged conductor to the uncharged conductor along the electric field. The potential will continually decrease along this path. A second path from the uncharged conductor to infinity will again continually lower the potential further. Therefore, we can say that the uncharged body must be intermediate in potential between that of the charged body and that of infinity.

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Question 453 Marks

Suppose, one wishes to construct a 1.0 farad capacitor using circular discs. If the separation between the discs be kept at 1.0mm, what would be the radius of the discs?

Answer

Let the radius of the disc = R

$\therefore$ Area $=\pi\text{R}^2$
$\text{C}=1\text{f}$
$\text{D}=1\text{mm}=10^{-3}\text{m}$
$\therefore\text{C}=\frac{\in_0\text{A}}{\text{d}}$
$\Rightarrow1=\frac{8.85\times10^{-12}\times\pi\text{r}^2}{10^{-3}}$
$\Rightarrow\text{r}^2=\frac{10^{-3}\times10^{12}}{8.85\times\pi}=\frac{10^9}{27.784}$
$=5998.5\text{m}=6\text{Km}$

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Question 463 Marks

Two capacitors of capacitances $4.0\mu\text{F}$ and $6.0\mu\text{F}$ are connected in series with a battery of 20V. Find the energy supplied by the battery.

Answer

$\therefore\text{C}_1=4\mu\text{F},\ \text{C}_2=6\mu\text{F},\ \text{V}=20\text{V}$
Eq. capacitor $\text{C}_{\text{eq}}=\frac{\text{C}_1\text{C}_2}{\text{C}_1+\text{C}_2}=\frac{4\times6}{4+6}=2.4$
$\therefore$ The Eq Capacitance $\text{C}_{\text{eq}}=2.5\mu\text{F}$
$\therefore$ The energy supplied by the battery to each plate
$\text{E}=\Big(\frac{1}{2}\Big)\text{CV}^2$
$=\Big(\frac{1}{2}\Big)\times2.4\times20^2=480\mu\text{J}$
$\therefore$ The energy supplies by the battery to capacitor $=2\times480=960\mu\text{J}$

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Question 473 Marks

Consider two conducting spheres of radii $R_1$ and $R_2$ with $R_1 > R_2$. If the two are at the same potential, the larger sphere has more charge than the smaller sphere. State whether the charge density of the smaller sphere is more or less than that of the larger one.

Answer

Since, the two spheres are at the same potential, therefore$\frac{1}{4\pi\in_0}\frac{\text{q}_1}{\text{R}_1}=\frac{1}{4\pi\in_0}\frac{\text{q}_2}{\text{R}_2}$
$\Rightarrow\ \frac{\text{R}_1}{\in_0}\frac{\text{q}}{4\pi\text{R}_1^2}=\frac{\text{R}_2}{\in_0}\frac{\text{q}_2}{4\pi\text{R}_2^2}$
or $\sigma_1\text{R}_1=\sigma_2\text{R}_2\Rightarrow\ \frac{\sigma_1}{\sigma_2}=\frac{\text{R}_2}{\text{R}_1}$ As $\text{R}_2>\text{R}_1$, this imply that $\sigma_1>\sigma_2$. The charge density of the smalller sphere is more than thet of the larger one.

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Question 483 Marks

Two charged particles, having equal charges of $2.0 \times 10^{-5}C$ each, are brought from infinity to within a separation of 10cm. Find the increase in the electric potential energy during the process.

Answer

$\text{q}_1=\text{q}_2=2\times10^{-5}\text{C}$Each are brought from infinity to 10cm a part $d = 10 \times 10^{-2}m$
So work done = negative of work done. (Potential E)
$\text{P.E}=\int\limits_\infty^{10}\text{F}\times\text{ds}$
$\text{P.E}=\text{K}\times\frac{\text{q}_1\text{q}_2}{\text{r}}$
$=\frac{9\times10^9\times4\times10^{10}}{10\times10^{-2}}$
$=36\text{J}$

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Question 493 Marks

Three point charges +Q, -2Q and -3Q are placed at the vertices of an equilateral triangle ABC of side l.
If these charges are displaced to the mid points $A_1, B_1$ and $C_1$ respectively, calculate the amount of work done in shifting the charges to the new locations.

Answer

Work done to put the three charges at A, B and C,

$\text{U}_{\text{initial}}=\frac{1}{4\pi\varepsilon_{0}}\Big[\frac{\text{Q}(-2\text{Q})}{\text{l}}+\frac{\text{Q}(-3\text{Q})}{\text{l}}+\frac{2\text{Q}(3\text{Q})}{\text{l}}\Big]$
$=\frac{1}{4\pi\varepsilon_0}\frac{1\text{Q}^2}{\text{l}}.$
Work done to take the three charges to position $A_1, B_1$ and $C_1,$
$\text{U}_{\text{initial}}=\frac{1}{4\pi\varepsilon_0}\bigg[\frac{\text{Q}(-2\text{Q})}{\frac{\text{l}}{2}}+\frac{\text{Q}(-3\text{A})}{\frac{\text{l}}{2}}+\frac{2\text{Q}(3\text{Q})}{\frac{\text{l}}{2}}\bigg]$
$=\frac{1}{4\pi\varepsilon_0}\frac{2\text{Q}^2}{\text{l}}.$
Work done to shift the charges to the new locations is:$\text{W}=\text{U}_{\text{final}}-\text{U}_{\text{initial}}=-\frac{1}{4\pi\varepsilon_0}\frac{2\text{Q}^2}{\text{l}}$

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Question 503 Marks

A metal sphere of radius R is charged to a potential V:

Find the electrostatic energy stored in the electric field within a concentric sphere of radius 2R.
Show that the electrostatic field energy stored outside the sphere of radius 2R equals that stored within it.

Answer

$\text{Q}=\text{CV}=4\pi\in_0\text{R}\times\text{V}$$\text{E}=\frac{1}2{}\frac{\text{q}^2}{\text{C}}$ $[\therefore$ 'C' in a spherical shell $=4\pi\in_0\text{R}]$
$\text{E}=\frac{1}{2}\frac{16\pi^2\in_0^2\times\text{R}^2\times\text{V}^2}{4\pi\in_0\times2\text{R}}=2\pi\in_0\text{RV}^2$ $[$'C' of bigger shell $=4\pi\in_0\text{R}]$

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