Question

The figure shows two identical parallel plate capacitors connected to a battery with the switch $S$ closed. The switch is now opened and the free space between the plates of the capacitors is filled with a dielectric of dielectric constant (or relative permittivity) $3$.Find the ratio of the total electrostatic energy stored in both capacitors before and after the introduction of the dielectric.

• A. $\frac{2}{3}$
• B. $\frac{5}{3}$
• C. $\frac{3}{5}$
• D. $\frac{3}{2}$

Answer 1

The correct option is C $\frac{3}{5}$

Energy in each capacitor is $\frac{1}{2}c{V}^2$

when switch is opened

$Q$ on $2nd$ capacitor is $cv$ even after introduction of dielectric $Q$ remain same but $C$ will become $KC$

Energy $=\frac{Q^2}{2C^2}=\frac{(cv{)}^2}{2\left(K_1\right)}=\frac{c{v}^2}{2k}$

Energy is set is $\frac{1}{2}c{v}^2$

same as before

after dielectric insertion

$c$ become $kc$

Energy $=\frac{1}{2}\left(kc\right)v^2$

$=\frac{kc{v}^2}{2}$

Total energy initially $=\frac{1}{2}c{v}^2+\frac{1}{2}c{v}^2=c{v}^2$------(1)

Total energy after opening switch $=k\frac{c{v}^2}{2}+\frac{c{v}^2}{2k}$

$=\frac{c{v}^2}{2}(k+\frac{1}{k})$

put $k=3$

$=\frac{c{v}^2\left(10\right)}{2\times 3}=\frac{5c{v}^2}{3}$-----(2)

(1) $÷$ (2)

$=3/5$