Question

An air capacitor with plates of area $1m^2$ and $0.01$ meter apart is charged with ${10}^{-6}$C of electricity. When the capacitor is submerged in oil of relative permittivity $2$, then the energy decreases by

• A. $20$ %
• B. $50$ %
• C. $60$ %
• D. $75$ %

Answer 1

The correct option is B $50$ %

given $A=1m^2;d=0\cdot 01$
we know $c=\frac{{\epsilon }_{0}A}{d}$
$energy{V}_1=\frac{1}{2}c{v}^{2.}$
when capacitance is submerged in oil c increases by k
$c_2=\frac{k{\epsilon }_{0}A}{d}=k{c}_1$
charge on the plates are constant
$\therefore V_2=\frac{V_1}{k}$
$V_2=\frac{1}{2}\cdot \left(kc\right)\cdot (\frac{V_1}{k}{)}^2=\frac{1}{2}kc\frac{V_1^2}{k}=\frac{1}{2}\frac{c{V}_1^2}{k}=\frac{1}{4}c{V}^2(k=2)$
$\therefore$ % change in energy $=\left(\frac{d\frac{1}{2}c{v}^2-\frac{1}{4}c{v}^2}{\frac{1}{2}c{v}^2}\right)\times 100=50$ %