Question

A capacitor is charged with a battery and then removed from the battery. In this specially designed capacitor, we are able to make the plate size (area) larger without changing anything else. If the plate area is made larger after the capacitor has been disconnected, what will happen to the charge on the plates, the voltage across the plates, and the value of capacitance for this capacitor?

Answer 1

Hint: In order to solve this problem, first we will find the dependence of charge, voltage, and capacitance for this capacitor by some basic formulae then we will substitute the value according to the values that got changed in the question. After that, we will find the changes in the quantities of charge, voltage, and capacitance when the area has been increased.

Complete step by step answer:
According to the question we have to use the formula for capacitance of a parallel plate capacitor $C = \dfrac{{A\varepsilon }}{d}$
Where, \[\varepsilon \]= Permittivity of the medium
$A$= Area of the plates
$d$= Distance between the plates
And also we know that charge in a capacitor is given by
$Q = CV$
Where
$Q$= Charge on the plates
$C$=Capacitance of the capacitor
$V$=Potential difference between the plates
Assume when the battery when connected to the specially designed capacitor the plates of the capacitor acquires a charge of $Q$ & the potential difference of $V_1$ is produced initially.
After that when the battery is removed then the capacitor will store the charge Q in its plate.
Let the area of the plates= $A_1$
Distance between the plates = $d$
Assume in this case capacitance of the capacitor = $C_1$
By using the relation
$C = \dfrac{{A\varepsilon }}{d}$
Substituting the values
${C_1} = \dfrac{{{A_1}\varepsilon }}{d}$ ………………….....(1)

We can see capacitance is proportional to the area of the plates.
When the Area of the plates is increased keeping the distance constant than by using the relation
Let the new area of the plates be $A_2$
Let the new capacitance between the plates be $C_2$
${C_2} = \dfrac{{{A_2}\varepsilon }}{d}$ …………………….…(2)
Where $C_2$ is the new capacitance & $A_2$ is the new area
According to question, we know that area has been increased i.e.
$\Rightarrow {A_1} < {A_2}$
Since ${A_1} < {A_2}$
therefore new capacitance of capacitor will be greater than initial capacitance from equation (1) & (2) i.e.
$\Rightarrow {C_1} < {C_2}$
We know that once the battery there will be no flow of charge hence the charge will remain constant.
Also using the relation we can say that
$Q = CV$
As charge in the plates does not change
We can substitute the values
\[\Rightarrow Q = {C_1}{V_1}\]
Let the new potential between the plates= V1
\[\Rightarrow Q = {C_2}{V_2}\]
\[\Rightarrow {C_1}{V_1} = {C_2}{V_2}\]
We can say that
$C\propto \dfrac{1}{V}$
Since ${C_1} < {C_2}$
$\therefore {V_1} < {V_2}$

So finally we conclude that:
1. Charge between the plates remains the same.
2. Voltage between the plates decreases after the area is increased.
3. Capacitance between the plates increases.

Note:
We must not forget that charge on the capacitor plates depends only on the external energy source i.e. battery with which it will be charged or any external circuit connected to it from which it will be discharged. In the question after charging no external circuit is connected so the charge will remain constant.