Question

A parallel-plate capacitor is connected to a battery. The space between the two plates is empty. If the separation between the capacitor plates is doubled while the capacitor remains connected to the battery, what is the ratio of the final stored energy to the initial stored energy?

Answer 1

Answer:

The final stored energy will become half.

Explanation:

 We know that stored energy in the capacitor is given as

$E=\dfrac{1}{2}CV^2$

C=capacitance

V=Voltage difference

E=Energy

$C=\dfrac{\varepsilon A}{d}$

d=Distance between plates

A=Area

$E=\dfrac{1}{2}\times \dfrac{\varepsilon A}{d}\times V^2$

If the distance between plates get double ,say d' = 2 d

Then stored energy

$E'=\dfrac{1}{2}\times \dfrac{\varepsilon A}{d'}\times V^2$

$E'=\dfrac{1}{2}\times \dfrac{\varepsilon A}{2d}\times V^2$

$E'=\dfrac {E}{2}$

Therefore the final stored energy will become half.