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.

  Capacitor Property	Value after Plate Area is increased
  Charge on Plates	smaller
  Voltage Across Plates	smaller
  Capacitance	smaller

• B.

  Capacitor Property	Value after Plate Area is increased
  Charge on Plates	larger
  Voltage Across Plates	smaller
  Capacitance	large

• C.

  Capacitor Property	Value after Plates Area is increased
  Change on Plates	same
  Voltage Across Plates	larger
  Capacitance	smaller

• D.

  Capacitor Property	Value after Plates Area is increased
  Change on Plates	Same
  Voltage Across Plates	Same
  Capacitance	larger

• E.

  Capacitor Property	Value after Plates Area is increased
  Change on Plates	same
  Voltage Across Plates	smaller
  Capacitance	larger

Answer 1

The correct option is E

Capacitor Property	Value after Plates Area is increased
Change on Plates	same
Voltage Across Plates	smaller
Capacitance	larger

Since the capacitor is charged first and then removed from the battery and then made changes to, the charge $Q$ on the capacitor remains the same.

The capacitance of the capacitor=$C=\frac{A{ϵ}_0}{d}$

Hence as the area of capacitor is increased, the capacitance of capacitor increases.

From the equation $Q=CV$

$⟹V\propto \frac{1}{C}$

Hence the voltage across the plates decreases.

Correct answer is option E.