Question

A parallel plate condenser with plate separation d is charged with the help of battery so that $V_0$ energy is stored in the system. The battery is now removed. a plate of dielectric constant k and thickness d is placed between the plates of condenser. The new energy of the system will be :

• A. $V_0{k}^{-2}$
• B. $k^2{V}_0$
• C. $V_0{k}^{-1}$
• D. $k{V}_0$

Answer 1

The correct option is C $V_0{k}^{-1}$

As the battery is removed from the capacitor so the charge (Q) remains constant.
If the capacitance of capacitor before inserting the dielectric is $C$, the capacitance of capacitor after inserting the dielectric is $kC$
Here, Energy $V_0=\frac{Q^2}{2C}$
After inserting dielectric the energy becomes $U=\frac{Q^2}{2kC}=V_0{k}^{-1}$