A capacitor of capacitance C is charged by a battery of emf V and then disconnected. The work done by an external agent to insert a dielectric of dielectric strength k of half the length of the capacitor is
Answer & explanation
Correct answer: option 2
Charge remains conserved
$q= C V\left(C_1=\frac{K C}{2}=\text { and } C_2=\frac{C}{2}\right)$
$q=\left(C_1+C_2\right) V' U_f-U_i=\frac{1}{2} \frac{C}{2}(1+k)\left(\frac{2 V}{1+k}\right)^2-\frac{1}{2} C V^2$
$\Rightarrow C V=\frac{C}{2}(k+1) V'$
$=\frac{C V^2}{1+k}-\frac{1}{2} C V^2$
$\Rightarrow V'=\frac{2 V}{k+1}=\frac{1}{2} C V^2\left(\frac{2}{1+k}-1\right)$
$=\frac{1}{2} C V^2\left(\frac{1-k}{1+k}\right)$.