3 Marks Question

Question 13 Marks

A parallel plate capacitor is charged to a potential difference V from a direct current source, while there is air between the plates. Without separating the capacitor from the battery, a dielectric medium of dielectric constant K has been filled in place of air. Explain with reasons which of the following changes will occur? Also give explanation of the answers.(i) Potential difference (ii) Electric field between the plates (iii) Capacitance (iv) charge (v) energy.

Answer

(i) Since here the capacitor is connected to the battery, i.e. the electric potential difference V will remain unchanged.
(ii) Since both the potential difference and the distance between the plates are unchanged, the electric field $E=\frac{V}{D}$ will remain unchanged.
(iii) Capacitance will increase $\because C = KC _0$
(iv) Since V is constant and the capacitance becomes K times, the charge Q will also become K times.
(v) Initial energy $U_0=\frac{1}{2} C_0\left(V_0\right)^2$ and final energy $U =\frac{1}{2} K C _0\left(V_0\right)^2= K U _0$ i.e.,energy will also increase K times.

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Question 23 Marks

A small conducting sphere (radius r ) is placed concentrically inside a larger conducting hollow sphere of radius $R$. The big and small spheres are charged with $Q$ and $q(Q>q)$ respectively and kept separate from each other. Calculate the potential difference between the spheres.

Answer

Potential on the surface of A due to Q $ V=\frac{1}{4 \pi \epsilon_0} \times \frac{Q}{R} $
Potential on the surface of A due to $q$

$ V=\frac{1}{4 \pi \epsilon_0} \times \frac{q}{R} $
Potential on the surface of $B$ due to $Q=\frac{1}{4 \pi \epsilon_0} \times \frac{Q}{R}$
Potential on the surface of B due to $q=\frac{1}{4 \pi \epsilon_0} \times \frac{q}{r}$
$\therefore$ Potential at $A , B _{s l}=\frac{1}{4 \pi \epsilon_0}\left(\frac{ Q }{ R }+\frac{q}{ R }\right)$
Potential at $B , V _{ B }=\frac{1}{4 \pi \epsilon_0}\left(\frac{ Q }{ R }+\frac{q}{r}\right)$
$\therefore \quad V _{ B }- V _{ A }=\frac{1}{4 \pi \epsilon_0}\left(\frac{q}{r}-\frac{q}{ R }\right)$

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Question 33 Marks

Write the product of equivalent capacitances when 10 capacitors, each having a capacitance of $10 \mu F$, are connected in series and then parallel connection.

Answer

Equivalent capacitance of series combination of $n$ capacitors will be
$ \frac{1}{C_{s}}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}+\ldots \ldots+\frac{1}{C_n} $
There are 10 capacitors each with a capacitance of $10 \mu F$.All of these are combined into series combinations. Therefore the value of equivalence will be $1 \mu F$.
All these have been combined in parallel combination after series combination.
Formula of equivalency for parallel combination :
$C _{ P }= C _1+ C _2+ C _3 \ldots \ldots C _n$
$\begin{array}{r}\therefore C _{ P }=10+10+10+10+10+10+10+10+ \\ 10+10=100 \mu F\end{array}$
The product of both the capacitances
$= C _{ S } \times C _{ P }=1 \mu F \times 100 \mu F$
$=1 \times 10^{-6} F \times 100 \times 10^{-6} F$
$=10^{-10} F=1 A F$ Ans.

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Question 43 Marks

What will be the effect on placing a metal foil between the plates of a parallel plate capacitor on its capacitance?

Answer

When there is air between the plates of the capacitor, capacitance $ C_0=\frac{\epsilon_0 A}{d} . $

By placing a metal foil between its plates $($Fig.$),$ this capacitor will be equivalent to a series combination of two parallel plate capacitors, in which the area of each plate is $A$ and the distance between the plates $=\frac{d}{2}$
Therefore, the capacitance of each of these is
$ C_1=\frac{E_0 A}{d / 2}=\frac{2 E_0 A}{d}$
$C_1=2 C_0 $
Therefore, the capacitance of their series combination
$ C=\frac{2 C_0 \times 2 C_0}{2 C_0+2 C_0}=C_0 $
Therefore, there will be no effect on the capacitance.

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Question 53 Marks

Define polarization.

Answer

The process of separating charges by pulling atoms in a dielectric in an induced electric field is called polarization. The pull of the atoms continues until the force exerted by the electric field on the charges becomes equal to the restoring force. Due to this stretching, the positive centre of gravity of the nucleus and the negative centre of gravity, which is formed due to the electrons, get separated from each other. And thus the atoms acquire dipole moment. That is, the dipole moment per unit volume of a substance is called its polarization and it is denoted by $\vec{p}$.
For linear isotropic dielectrics $ \vec{p}=\chi_0 E $
Here, $\chi_0$ is a constant characteristic of the dielectric which is called the electrical tendency of the dielectric medium.

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Question 63 Marks

In a parallel plate capacitor with air between the plates, each plate has an area of 6 x 10^-3m² 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 parallel plate capacitor A = 6 x 10^-3 m²
Distance between the plates d = 3 mm
$=3 \times 10^{-3} m$
$\in_0=8.854 \times 10^{-12} C ^2 N^2 m^{-2}$
Let the capacitance of a parallel plate capacitor with air between the plates = C₀Hence using the formula $C _0=\frac{\in_0 A}{d}$
On putting values $\quad C _0=\frac{8.854 \times 10^{-12} \times 6 \times 10^{-3}}{3 \times 10^{-3}}$
$=17.708 \times 10^{-12} F$
$=17.708 p F$
$=18 pF$
Supply applied on the capacitor $= V _0=100 V$.
Let the value of charge on each plate of the capacitor
$= Q _0=?$
Using the formula $Q_0=C_0 V_0$
$=17.708 \times 10^{-12} \times 100$
$=17.708 \times 10^{-10} C$
$=1.7708 \times 10^{-9} C$
$=1.7708 nC$
$\cong 1.8 nC$

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Question 73 Marks

Three capacitors of capacitance 2 pF, 3 pF and 4 pF are connected in parallel.
(a) What is the total capacitance of the combination?
(b) Determine the charge on each capacitor if the combination is connected to a 100 V supply.
C₁ = 2 pF
= 2 X 10 ^-12F
C₂ = 3 pF
3 X 10 ^-12 F
C₃ = 4 pF
= 4 X 10^-12 F

Answer

Given:
Given potential of the combination = V

V = 100 V
(a) In parallel sequence, from the formula
$C _{ p }= C _1+ C _2+ C _3$
$C _{ p }=2 pF +3 pF +4 pF$
$=9 pF =9 \times 10^{-12} F$
(b) Let the charges on each capacitor be Q1, Q 2 and Q3 respectively.
From the formula $Q=C V$ and $V=100$ volts
$Q _1= C _1 V=2 \times 10^{-12} \times 100=2 \times 10^{-10} C$
$Q_2=C_2 V=3 \times 10^{-12} \times 100=3 \times 10^{-10} C$
$Q _3= C _3 V=4 \times 10^{-12} \times 100=2 \times 10^{-10} C$

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Question 83 Marks

A parallel plate capacitor with air between its plates has a capacitance of 8 pF(1 pF = 10^-12F) . 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

(i) Here is a parallel plate capacitor, which has air between its plates.
$C_1=\frac{\in_0 A}{d}=8 pF =8 \times 10^{-12} F$
(ii) When a substance of dielectric constant 6 is filled in the space between the plates
$C _2= K \frac{\in_0 A}{d^{\prime}}$
Here it is given K = 6 d' = d / 2
$C _2=\frac{6\left(\in_0 A\right)}{d / 2}=2 \times 6 \times\left(\frac{\in_0 A}{d}\right)$
$C_2=12 C_1$
but $C_1=8 \times 10^{-12} F$
Hence on putting values
$C _2=12 \times 8 \times 10^{-12} F$
$=96 \times 10^{-12} F=96 pF$

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Question 93 Marks

Two charges 2 $\mu C$ and - 2$\mu C$ are placed at points A and B 6 cm apart.
(a) Identify an equipotential surface of the system.
(b) What is the direction of the electric field at every point on this surface?

Answer

(a) It is given that two charges 2 µC and - 2 $\mu C$are placed on A and B.

AB = 6 cm = 6 x 10^-2m
The equipotential surface of a system of two given charges is normal to the line joining A and B. The surface passes through the midpoint C of AB. potential at point C
$\frac{1}{4 \pi \in_0} \frac{\left(2 \times 10^{-6}\right)}{3 \times 10^{-2}}+\frac{1}{4 \pi \in_0} \frac{\left(-2 \times 10^{-6}\right)}{3 \times 10^{-2}}$
That means the potential at point C will be zero.
Thus, the potential at every point on this surface is equal and it is zero, hence it is an equipotential surface.
(b) We know that electric field always acts from + to-charge. Thus, here the electric field works from positively charged (+ve) point A towards negatively charged (-ve) point B and it is in the AB direction normal to the equipotential surface.

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Question 103 Marks

A regular hexagon of side 10 cm has a charge of 5 $\mu C$ at each of its vertex. Calculate the potential at the centre of the hexagon.

Answer

ABCDEF is a regular hexagon, each side of which = 10cm .
The center of the hexagon is O,

From geometry
OA = OB = OC = OD = OE = OF = 10 cm .
Charge q on A, B, C, D, E and F = 5 $\mu C$
Electric potential on O due to all charges
$V =6 \times \frac{1}{4 \pi \in_0} \frac{q}{r}$
Given : $r=10 cm=10 \times 10^{-2} m$ and
$q=5 \mu C =5 \times 10^{-6} C$
On putting values
$V=\frac{6 \times 9 \times 10^9 \times 5 \times 10^{-6}}{10 \times 10^{-2}}$
$V=27 \times 10^{11-6}=27 \times 10^5$
or $V=2.7 \times 10^{6} V$

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Physics 3 Marks Question

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