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

$\frac{1}{2} CV^2\left(\frac{1-k}{k+1}\right)$

Charge remains conserved

$q= C V\left(C_1=\frac{K C}{2}=\text { and } C_2=\frac{C}{2}\right)$

$q=\left(C_1+C_2\right) V' U_f-U_i=\frac{1}{2} \frac{C}{2}(1+k)\left(\frac{2 V}{1+k}\right)^2-\frac{1}{2} C V^2$

$\Rightarrow C V=\frac{C}{2}(k+1) V'$

$=\frac{C V^2}{1+k}-\frac{1}{2} C V^2$

$\Rightarrow V'=\frac{2 V}{k+1}=\frac{1}{2} C V^2\left(\frac{2}{1+k}-1\right)$

$=\frac{1}{2} C V^2\left(\frac{1-k}{1+k}\right)$.