The energy of a parallel plate capacitor when connected to a battery is $E$ . With the battery still in connection, if the plates of the capacitor are separated so that the distance between them is twice the original distance, then electrostatic energy becomes :

Question

The energy of a parallel plate capacitor when connected to a battery is

E

. With the battery still in connection, if the plates of the capacitor are separated so that the distance between them is twice the original distance, then electrostatic energy becomes :

Toppr Admin · Accepted Answer

-E-12Cv2-xA0-xA0-xA0-xA0-x21D2-C-x221D-1d-x21D2-C1C2-d2d1-2dd-2-x21D2-E1E2-C1C2-21-x21D2-EE2-21-x21D2-E2-E2

The energy of a parallel plate capacitor when connected to a battery is E. With the battery still in connection, if the plates of the capacitor are separated so that the distance between them is twice the original distance, then electrostatic energy becomes :E22E4EE4

E=12Cv2 ⇒C∝1d⇒C1C2=d2d1=2dd=2 ⇒E1E2=C1C2=21 ⇒EE2=21 ⇒E2=E2

$E = \frac{1}{2} C v^{2}$
$\Rightarrow C \propto \frac{1}{d}$
$\Rightarrow \frac{C_{1}}{C_{2}} = \frac{d_{2}}{d_{1}} = \frac{2 d}{d} = 2$

$\Rightarrow \frac{E_{1}}{E_{2}} = \frac{C_{1}}{C_{2}} = \frac{2}{1}$

$\Rightarrow \frac{E}{E_{2}} = \frac{2}{1}$

$\Rightarrow E_{2} = \frac{E}{2}$