Question

A parallel plate capacitor of capacitance C is charged and disconnected from the battery. The energy stored in it is E. If a dielectric slab of dielectric constant 6 is inserted between the plates of the capacitor then energy and capacitance will become :

• A. E, C
• B. 6E, 6C
• C. $\frac{E}{6},6C$
• D. E, 6C

Answer 1

Answer: (C)

Let C be charged to v
$\Rightarrow E=\frac{1}{2}\times C\times V^2$
Now when dielectric is added.
C increases by kC and as battery is disconnected $\upsilon$ is constant.
$\therefore$ V decreases by k.
$\therefore C_1=kC;V_1=\frac{V}{k}$
$\Rightarrow E_1=\frac{1}{2}{c}_1{v}_1^2$
$E_1=\frac{1}{2}kc\times \frac{v}{k}\times \frac{v}{k}$
$E_1=\frac{1}{2}c\frac{v^2}{6}$
$c_1=\frac{1}{6}\cdot E$
$\therefore E_1=\frac{E}{6}$
and $C_1=6C$