Question

Two identical sheets of a metallic foil are separated by d and capacitance of the system is C and charged to a potential difference E. Keeping the charge constant, the separation is increased by l. Then the new capacitance and the potential difference will be:
A) $\dfrac{{{\varepsilon _0}A}}{d}.E$
B) $\dfrac{{{\varepsilon _0}A}}{{(d + l)}}.E$
C) $\dfrac{{{\varepsilon _0}A}}{{(d + l)}},(1 + \dfrac{l}{d})E$
D) $\dfrac{{{\varepsilon _0}A}}{d},(1 + \dfrac{l}{d})E$

Answer 1

Hint: A capacitor is the device that stores energy in the form of an electric charge. It consists of two electrical conductors separated by a distance and the space between them is filled with an insulating material or dielectric. The ability of the capacitor to store charge is known as capacitance.

Complete step by step solution:
Step I: The capacitance is the ratio of the change in electric charge of the system to the change in electric potential.
It is written as
$C = \dfrac{Q}{V}$---(i)
Where C is the capacitance
V is the potential and
Q is the charge

Step II: From equation (i) it can be written that the capacitance of the one of the metallic foil is $Q = CV$---(ii)
Since the charge will remain constant, the capacitance of the other metallic foil is written as
$ \Rightarrow Q = {C_1}{V_1}$---(iii)
Comparing equations (ii) and (iii),
$ \Rightarrow Q = CV = {C_1}{V_1}$---(iv)

Step III: The value of C for one of the metallic foil is
$ \Rightarrow C = \dfrac{{{\varepsilon _0}A}}{d}$
For another metallic foil, the charge is constant and the separation is increased by a distance l, so the new capacitance is given by
$ \Rightarrow {C_1} = \dfrac{{{\varepsilon _0}A}}{{(d + l)}}$---(v)
Substituting the value of capacitance in equation (iv),
$ \Rightarrow Q = \dfrac{{{\varepsilon _0}AE}}{d} = \dfrac{{{\varepsilon _0}A{E_1}}}{{d + l}}$---(vi)

Step IV: From equation (v) it can be written that
$\dfrac{{{\varepsilon _0}AE}}{d} = \dfrac{{{\varepsilon _0}A{E_1}}}{{d + l}}$
$ \Rightarrow {E_1} = \dfrac{{d + l}}{d}E$
$ \Rightarrow {E_1} = (1 + \dfrac{l}{d})E$---(vii)
Therefore, the new capacitance will be $\dfrac{{{\varepsilon _0}A}}{{d + l}}$and the new potential difference will be $(1 + \dfrac{l}{d})E$.

Option C is the right answer.

Note: It is important to note that two metallic sheets will behave like a capacitor. When a potential difference is applied across the capacitor one plate of the capacitor will have positive charge and the other will have a negative charge. In the steady state, the current will flow from the positive terminal of the capacitor to the negative terminal.