A parallel plate capacitor filled with a medium of dielectric constant $10$ , is connected across a battery and is charged. The dielectric slab is replaced by another slab of dielectric constant $15$ . Then the energy of capacitor will ......................
JEE MAIN 2022, Diffcult
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$E \Rightarrow \frac{1}{2}( KC ) v ^{2}$
$\therefore \%$ change
$\Rightarrow \frac{\frac{1}{2} K _{2} CV ^{2}-\frac{1}{2} K _{1} CV ^{2}}{\frac{1}{2} K _{1} CV ^{2}}=\frac{ K _{2}- K _{1}}{ K _{1}} \times 100$
$\Rightarrow \frac{15-10}{10} \times 100=50 \%$
$\therefore \%$ change
$\Rightarrow \frac{\frac{1}{2} K _{2} CV ^{2}-\frac{1}{2} K _{1} CV ^{2}}{\frac{1}{2} K _{1} CV ^{2}}=\frac{ K _{2}- K _{1}}{ K _{1}} \times 100$
$\Rightarrow \frac{15-10}{10} \times 100=50 \%$
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