Plates of a parallel plate capacitor $(C = 6 n F)$ carries charge $Q_{1} = 12 n C$ and $Q_{2} = 6 n C$ , then energy stored in capacitor is

\frac{4}{5} n J

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{3}{4} n J

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

\frac{2}{3} n J

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{1}{3} n J

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $\frac{3}{4} n J$

At equilibrium,

$E$ inside the conductor with charge $Q_{1}$ is $= 0$ . So,

$\frac{Q_{1} - x}{2 A ϵ_{0}} - \frac{x}{2 A ϵ_{0}} + \frac{x}{2 A ϵ_{0}} - (\frac{Q_{2} + x}{2 A ϵ_{0}}) = 0$

$Q_{1} - x - Q_{2} - x = 0$

$x = \frac{Q_{1} - Q_{2}}{2}$

$Δ V$ between plates $= \frac{x}{C} = \frac{Q_{1} - Q_{2}}{2 C}$
$= \frac{12 - 6}{2 \times 6} = 0.5 v o l t$
$V = \frac{1}{2} . C Δ V^{2} = \frac{1}{2} \times 6 \times (\frac{1}{2})^{2} = \frac{6}{8} = \frac{3}{4} n J$

Suggest Corrections

Similar questions

Q. Plates of a parallel plate capacitor

(C = 6 n F)

carries charge

Q_{1} = 12 n C

and

Q_{2} = 6 n C

, then energy stored in capacitor is

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 :

Q. Prove that the energy stored in a parallel plate capacitor is