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?
A) All decreases.
B) Decreases, Increases and Increases, respectively.
C) Constant, Increases and Decreases, respectively.
D) Constant, Constant and Increases, respectively.
E) Constant, Decreases and Increases, respectively.

Answer 1

Hint:- This problem is related to the capacitors, here after charging the plates of the capacitors are changed, so some of the properties are changed. The property of the capacitor change can be determined by using two formulas, the capacitance of the capacitor formula and the charge of the capacitor formula. By using these two formulae, the solution can be determined.

Useful formula:
The capacitance of the capacitor,
$C = \dfrac{{A{\varepsilon _0}}}{d}$
Where, $C$ is the capacitance, $A$ is the area of the plates, ${\varepsilon _0}$ is the permittivity and $d$ is the distance between two plates.
The charge of the capacitor,
$Q = CV$
Where, $Q$ is the charge, $C$ is the capacitance and $V$ is the voltage

Complete step by step solution:
Since, the capacitor is first charged and the charger is removed from the battery and then the plates of the capacitor are changed..
By this formula,
$Q = CV$,
The charge of the capacitor is not related to the area of the plates. So, if the plates of the capacitor are changed, the charge of the capacitor remains the same.
By this formula,
$C = \dfrac{{A{\varepsilon _0}}}{d}$
If the area of the capacitor is increased, then the capacitance is increased.
By this formula,
$Q = CV$ or $V \propto \dfrac{1}{C}$
The voltage and the capacitance are inversely proportional, by changing the area, the capacitance is increased, then the voltage is decreased.

Hence, the option (E) is correct.

Note: By the two formulas, the capacitance of the capacitor and the charge of the capacitor, then the solution can be determined. In the charge of the capacitor formula, the charge $Q$ is constant, so if the capacitance $C$ comes to the other side, it will go to the denominator. Then, the voltage and capacitance are inversely proportional.