A slab of material of dielectric constant k, having the same area as the plate of a parallel plate capacitor but has a thickness $\frac{3}{5}$ of the distance between the plates d, is inserted between the plates of the capacitor. When there is air between the plates, the potential difference is $V_0$. What is the potential difference between the plates of the capacitor after inserting the dielectric?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → $\left(\frac{2 k+3}{5 k}\right) V_0$
Capacitance without Di-electric, $C_0=\frac{ε_0}{d}$
Now,
$C_1$ (Capacitance with Dielectric) = $\frac{Kε_0A}{d/3}=3.\frac{Kε_0A}{d}$
$C_2$ (Capacitance without Dielectric) = $\frac{Kε_0A}{2d/3}=\frac{3ε_0A}{2d}$
$∴\frac{1}{C}=\frac{1}{C_1}+\frac{1}{C_2}$ [Connected in series]
$\frac{1}{C}=\frac{d}{3Kε_0A}+\frac{2d}{3ε_0A}$
$⇒C=\frac{3ε_0A}{d\left(\frac{1}{k}+2\right)}$
Also, $V=\frac{Q}{C}=\frac{C_0V_0}{C}=\frac{\frac{ε_0}{d}V_0}{\frac{3ε_0A}{d\left(\frac{1}{k}+2\right)}}$
$=\frac{V_0}{3\left(\frac{1}{k}+2\right)}$