Question

A parallel plate capacitor filled with a dielectric relative permittivity $5$ between its plates is charged to acquire an energy E and isolated. If the dielectric is replaced by another of relative permittivity $2$, its energy becomes :

• A. $E$
• B. $0.4E$
• C. $2.5E$
• D. $6.25E$

Answer 1

Answer: (C)

$2.5E$

Let $C_0$ be the capacitance of air filled capacitor.

When the dielectric material with dielectric constant K is filled between the plates, the capacitance becomes, $C^{\prime }=K{C}_0$

As the capacitor is isolated, so total charged is conserved.

Thus, energy stored in capacitor is $E=\frac{q^2}{2C}$

As charged is conserved so $E\propto 1/C$

Thus, $E_2/E_5=5C_0/2C_0=2.5$