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Physics 5 Marks Questions

Electrostatic Potential and Capacitance · Rajasthan Board (RBSE) STD 12 Science (Rajasthan - English Medium). 19 questions.

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Question 15 Marks
What is capacitance of a conductor? What is its unit? Find the formula for capacitance of an isolated spherical conductor. Draw a graph between capacitance and radius. If a spherical conductor is placed in dielectric constant $\epsilon_r$, then what will be the effect on the value of capacitance?###If the area of each conducting plate of a parallel plate capacitor is A and the separation between them is $d$, then express the formula for its capacitance.
Answer
In the figure two plates $P _1$ and $P _2$, are of the same size. Both the plates are fixed to two insulated stands parallel to each other. The medium between the plates is air or vacuum. In the figure, the $P _2$ plate is earthed.
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Let the area of each plate be A and the distance between them be $d$. By giving $+Q$ charge to plate $P_1$, due to induction $-Q$ charge is produced on the inner surface of plate $P_2$, and $+Q$ charge is produced on the outer surface. Since the outer surface of plate $P_2$ is connected to the earth, its positive charge goes into the earth.
Suppose the surface density of charge on each plate is $\sigma$, therefore the electric field at any point between the plates is
$ E=\frac{\sigma}{\varepsilon_0} $
Here $\epsilon_0$ is the electrical conductivity of vacuum. Area of each plate A and charge is Q.
Therefore, $ \sigma=\frac{Q}{A} $
On putting the value of $\sigma$ from equation (2) into equation (1)
$ E=\frac{Q}{\varepsilon_0 A} $
If the potential difference between the plates due to this electric field is V then :
$V = E \times d$
$V =\frac{ Q d}{\varepsilon_0 A}$
If the capacitance C of a parallel plate capacitor is : $ C=\frac{Q}{V} $
On putting the value of V from equation (4)
$C =\frac{ Q }{\frac{ Q d}{\epsilon_0 A}}=\frac{\varepsilon_0 A}{d}$
$C =\frac{ Q }{\frac{ Q d}{\epsilon_0 A}}=\frac{\varepsilon_0 A}{d}$
$C =\frac{\varepsilon_0 A}{d}$ 
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Question 25 Marks
Write the definition of electric capacitance. A parallel plate capacitor is filled with a medium of dielectric constant K between the plates. Find the expression of its capacitance. Draw the necessary diagram.
Answer
When an isolated conductor is charged, its potential increases. The increase in potential is proportional to the charge supplied to the conductor.
$Q \propto V$ or $Q=C V$
Here the proportionality constant C is called the capacitance of the conductor. The value of capacitance C of a conductor depends on its size, shape and the medium used in its construction. Keep in mind that the value of capacitance of a conductor does not depend on the charge on the conductor and the increase in potential across it. If V $=1$ volt in equation (1), then $Q = C$ i.e., the capacitance of the conductor is equivalent to the amount of charge which increases the potential of the conductor by unit amount.
The S.I. unit of capacitance is coulomb/volt, which is called Farad. If in equation (1) $Q =1$ coulomb, $V =1$ volt then $C =1$ Farad.i.e., if on giving 1 coulomb of charge to a conductor, its potential increases by 1 volt, then the capacitance of that conductor is called 1 Farad.
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Parallel Plate Capacitor Completely Filled with a Dielectric Here the picture shows a parallel plate capacitor completely filled with dielectric material. The dielectric constant of the material is K and the area of the plates of the capacitor is A and the distance between them is $d$. If in the absence of dielectric material, the charge on the capacitor is $Q _0$ and the potential difference between the plates is $V _0$, then the capacitance of the capacitor in air or vacuum will be $ C_0=\frac{Q_0}{V_0} $
When a material of relative dielectric constant K is filled between the plates of a capacitor, the value of charge on the capacitor does not change due to the presence of the dielectric material filled between the plates. Rather, the value of the potential difference between the plates decrease by $\left(\frac{1}{K}\right)$ times. Therefore potential difference between the plates
$ V=\frac{V_0}{K} $
If the value of capacitance of a capacitor filled with dielectric constant is C then
$C=\frac{Q_0}{V}=\frac{Q_0}{V_0 / K}$
$C =\frac{ Q _0 K}{ V _0}$
$C = C _0 K$
or $C = K \left(\frac{\epsilon_0 A}{d}\right)$
Therefore, the value of capacitance C of a parallel plate capacitor completely filled with dielectric material is K times the value of air or vacuum capacitance $C_0$.
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Question 35 Marks
Obtain the relation for equivalence in series combination of capacitors. Make a circuit diagram.
Answer
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Question 45 Marks
Define capacitor. Obtain the relation of equivalence in series combination of capacitors by making a circuit diagram. Write the value of voltage $V_1$ in the given circuit.
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Answer
SELF
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Question 55 Marks
Define common potential. Find the expression for redistribution of charges and energy loss when capacitors are connected together.
Answer
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Question 65 Marks
Prove that the value of energy stored in a capacitor is $\frac{1 Q^2}{2 C}$ and the energy density of the electric field is $u=\frac{1}{2} \epsilon_0 E ^2$.
Answer
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Question 75 Marks
Establish the formula for the capacitance of a parallel plate capacitor in the presence of dielectric material completely and partially.
Answer
SELF
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Question 85 Marks
Define the capacitance of a capacitor. Obtain an expression for the capacitance of a parallel plate capacitor. And explain on what factors the does capacitance depend?
Answer
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Question 95 Marks
What is called a capacitor? Explain its principle. Find the formula for combination of capacitors in parallel and series.###Define capacitor. Establish the formula of resultant capacitance in its series and parallel combination. Also draw the necessary circuit diagram.###Define capacitor. How can we increase its capacity? Establish the formula for combining capacitors in parallel and series.
Answer
SELF
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Question 105 Marks
Calculate the work done in moving a test charge q0 from point A to point B against the repulsive force acting on it due to the other charge Q > 0 located at the origin.
Answer
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Question 125 Marks
Describe the potential energy of an electric dipole in an external field in different positions also draw the diagram.
Answer
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Question 135 Marks
Define electric dipole moment. Calculate the value of electric potential generated from electric dipole and prove that its value is maximum at the point situated on the axis and zero at the equatorial point? ###Define electric dipole moment of electric dipole. Obtain an expression for the electric potential at a common point due to electric dipole. Also draw necessary diagram.
Answer
SELF
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Question 145 Marks
What is meant by electric potential? Calculate the electric potential generated by a point charge. Draw the equipotential surfaces resulting from this.
Answer
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Question 155 Marks
Define electric potential difference and prove that the electric potential at any point inside the electric field is equal to the ratio of the work W done in bringing a test charge from infinity to that point and the value of the test charge $q_0$.
Answer
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Question 165 Marks
Explain what would happn if in the capacitor given in Q. 2.8, a 3 mm thick mica sheet (of dielectric constant = 6) were inserted between the plates.
(a) while the voltage supply remained connectd.
(b) after the supply was disconncted.
Answer
Whenever we place any insulating medium (dielectric constant K) between the plates of a capacitor, the capacitance of the capacitor will increase by K times. (a) If the potential supply (battery) remains connected to the capacitor then the potential will not change. Due to increase in capacitance the capacitor will acquire more charge (K times) from the battery. (b) If the battery is removed from the circuit, the charge of the battery will remain unchanged, but the potential will remain Kth part (V = Q / C)
(a) Let the capacitance of the capacitor with air medium = c0,
Dielectric constant K = 6
Therefore, the capacitance of the capacitor is
$C = KC _{ o }$
$=6 \times 18 p F$
$=108 \times 10^{-12} F=108 p F$
In this way, after placing the mica strip, the capacitance of the capacitor increases by K (=6) .
Potential on capacitor = 100 V
[Because the battery is connected to the capacitor.]
Therefore, charge $Q^{\prime}=C V$
$=108 \times 10^{-12} \times 100 V$
$=108 \times 10^{-10}$
$=1.08 \times 10^{-8} C$
Obviously, the charge on the plates becomes K times of the charge on the air medium, that is, when the supply remains connected and the mica strip is the medium, the charge increases.
(b) Capacitance of capacitor with mica medium
$C = KC _0=108 \times 10^{-12} F$
$=108 p F$
When the supply is removed, there will be no change in charge and the value of changed potential will be
$V^{\prime}=\frac{Q}{C^{\prime}}=\frac{Q}{K C}=\frac{V}{K}$
that is, now the potential will remain $K ^{\text {th }}$ part,
$V^{\prime}=\frac{100}{6}=16.67 V$
That is, the charge on the capacitor will remain the same with the mica medium as with the air medium.
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Question 175 Marks
A spherical conductor of radius 12cm has a charg of $1.6 \times 10^{-7} C$ distributed uniformly on its surface. What is the electric field:
(a) inside the sphere
(b) just outside the sphere
(c) at a point 18 cm from the centre of the sphere?
Answer
Given: Charge on conductor
$q=1.6 \times 10^{-7} C$
and the radius of the spherical conductor
$\begin{aligned} r & =12 cm
\\& =12 \times 10^{-2} m\end{aligned}$
(a) We know that the charge given to a spherical conductor remains on its surface.
Therefore electric field inside a spherical conductor is zero.
$\because \quad \phi=\oint_{ S } \overrightarrow{ E } \cdot \overrightarrow{d s}=\frac{q}{\in_0}=0$
( $\because$ Here $q=0$ Inside the conductor)
$\begin{aligned} \text { Or } & \overrightarrow{ E } \cdot \overrightarrow{ ds } & =0 \\ \text { Or } &
E & =0\end{aligned}$
(b) Just outside the sphere: The charge at a point just outside the sphere, i.e. at a point on the surface, can be considered concentrated at the center of the sphere. Thus
$E=\frac{1}{4 \pi \in_0} \frac{9}{R^2}$ Using the relationship
$=\frac{9 \times 10^9 \times 1.6 \times 10^{-7}}{\left(12 \times 10^{-2}\right)^2}=\frac{9 \times 1.6 \times 10^2 \times 10^4}{144}$
$=10^5 N / C$
(c) Position at 18 cm from the center of the sphere r = 18cm = 18 x 10-2 m
$E =\frac{1}{4 \pi \in_0} \frac{q}{r^2}$
on putting values
$=\frac{9 \times 10^9 \times 1.6 \times 10^{-7}}{\left(18 \times 10^{-2}\right)^2}$
$=\frac{14.4 \times 10^2}{324 \times 10^{-4}}=4.44 \times 10^4 N / C$
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Question 185 Marks
A 600 pF capacitor is charged by a 200 V supply. It is then disconnected from the supply and is connected to another uncharged 600 pF capacitor. How much electrostatic enery is lost in the process?
Answer
Given: $C _1=600 pF =600 \times 10^{-12} F$
$=6 \times 10^{-10} F$
$V _1=200 Volt$
Initial electric energy $U_1=\frac{1}{2} C_1 V_1^2$
On putting value $\quad U _1=\frac{1}{2} \times 6 \times 10^{-10} \times(200)^2$
$=\frac{1}{2} \times 6 \times 4 \times 10^{-10} \times 10^4$
$U _1=12 \times 10^{-6} J$
$C_2=6 \times 10^{-10} F$
$V _2=0$ (capacitor is uncharged)
If V is common potential then
$V=\frac{C_1 V_1+C_2 V_2}{C_1+C_2}$
$=\frac{6 \times 10^{-10} \times 200+6 \times 10^{-10} \times 0}{6 \times 10^{-10}+6 \times 10^{-10}}$
$=\frac{12 \times 10^{-8}}{12 \times 10^{-10}}=100 Volt$
Final electric energy
$U _2=\frac{1}{2}\left( C _1+ C _2\right) V ^2$
On putting values
$U _2=\frac{1}{2} \times 12 \times 10^{-10} \times(100)^2$
$\begin{array}{r}\because C _1+ C _2=6 \times 10^{-10} F+6 \times 10^{-10} \\ =12 \times 10^{-10} F\end{array}$
$=6 \times 10^{-10} \times 10^4=6 \times 10^{-6} J$
Loss in potential energy
$\Delta U=U_1-U_2$
$=12 \times 10^{-6}-6 \times 10^{-6}$
$=6 \times 10^{-6} J$
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Question 195 Marks
Two charges of$5 \times 10^{-8} C$ and $-3 \times 10^{-8}$ are located at a distance of 16 cm apart. At what point on the line joining the two charges is the electric potential zero? Take the potential at infinity to be zero.
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Answer
Given :
$q_1=5 \times 10^{-8} C$,
$q_2=-3 \times 10^{-8} C$
and $\quad r=16 cm=0.16 m$
Let the value of potential at point P located at a distance x from be zero. For this reason, distance of P $q_2=(0.16-x) m$,from potential at point P due to $q_1$ will be
$V _1=\frac{1}{4 \pi \in_0} \frac{q_1}{x}$
The value of electric potential due to $q_2$ at point P will be
$V _2=\frac{1}{4 \pi \in_0} \frac{q_1}{(0.16-x)}$
But the value of total potential at point $P$ is zero. Then $V_1+V_2=0$
or$\quad$$\frac{1}{4 \pi \in_0} \frac{q_1}{x}+\frac{1}{4 \pi \in_0} \frac{q_2}{(0.16-x)}=0$
or $\quad$$\frac{1}{4 \pi \in_0} \frac{q_1}{x}=\frac{1}{4 \pi \in_0} \frac{-q_2}{(0.16-x)}$
or $\quad$$\frac{q_1}{-q_2}=\frac{x}{(0.16-x)}$
$\frac{q_2}{x}=\frac{-q_2}{(0.16-x)}$
On putting values, $\frac{5 \times 10^{-8}}{3 \times 10^{-8}}=\frac{x}{(0.16-x)}$
or$\quad$$\frac{5}{3}=\frac{x}{(0.16-x)}$
or $\quad$$0.8-5 x=3 x$
$\Rightarrow \quad 0.8=8 x$
$\Rightarrow \quad x=\frac{0.8}{8}=0.1 m$
That is $\quad x=10 cm$
Now we will find another point where the value of potential will be zero. This point may be possible on the right side of the negative charge. If the distance of this point from the positive charge is x m then its distance from the negative charge will be (X - 0.16 ) m. Then $V_1+V_2=0$
then $\frac{1}{4 \pi \in_0} \frac{q_1}{x}+\frac{1}{4 \pi \in_0} \frac{q_2}{x-0,16}=0$
or$\quad$$\frac{q_1}{x}+\frac{q_2}{(x-0.16)}=0$
or$\quad$$\frac{q_1}{-q_2}=\frac{x}{(x-0.16)}$
on putting values
$\frac{5 \times 10^{-8}}{3 \times 10^{-8}}=\frac{x}{(x-0.16)}$
$\Rightarrow \quad \frac{5}{3}=\frac{x}{(x-0.16)}$
$\Rightarrow \quad 5 x-0.8=3 x$
$\Rightarrow \quad 5 x-3 x=0.8$
$\Rightarrow \quad 2 x=0.8$
$\Rightarrow \quad x=\frac{0.8}{2}=0.4 m$
or$\quad$ $x=40 cm$.
Therefore the value of potential will be zero. 10 cm, 40 cm away from positive charge towards negative charge.
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