A parallel plate capacitor is charged and then disconnected from the charging battery. If the plates are now moved farther apart by pulling at them by means of insulating handles, then

Question

A parallel plate capacitor is charged and then disconnected from the charging battery. If the plates are now moved farther apart by pulling at them by means of insulating handles, then (A) the energy stored in the capacitor decreases (B) the capacitance of the capacitor increases (C) the charge on the capacitor decreases (D) the voltage across the capacitor increases

Tardigrade Admin · Accepted Answer

We know that
The parallel plate capacitor

C = \frac{ε _{0} \cdot A}{d}

When

d

(distance between two parallel plate) increase, then

C

will be decrease

∵ Q = C V

where

Q =

constant

∴

The voltage across the capacitor increases.

Q. A parallel plate capacitor is charged and then disconnected from the charging battery. If the plates are now moved farther apart by pulling at them by means of insulating handles, then

the energy stored in the capacitor decreases

the capacitance of the capacitor increases

the charge on the capacitor decreases

the voltage across the capacitor increases

We know that The parallel plate capacitor C=dε0​⋅A​ When d (distance between two parallel plate) increase, then C will be decrease ∵Q=CV where Q= constant ∴ The voltage across the capacitor increases.

We know that
The parallel plate capacitor $C = \frac{ε _{0} \cdot A}{d}$
When $d$ (distance between two parallel plate) increase, then $C$ will be decrease
$∵ Q = C V$ where $Q =$ constant
$∴$ The voltage across the capacitor increases.