Q. Between the plates of a parallel plate condenser, a plate of thickness $t_{1}$ and dielectric constant $k_{1}$ is placed. In the rest of the space, there is another plate of thickness $t_{2}$ and dielectric constant $k_{2}.$ The potential difference across the condenser will be
A) $\frac{Q}{A\varepsilon_{0}}\left( \frac{t_{1}}{k_{1}} + \frac{t_{2}}{k_{2}} \right)$
B) $\frac{\varepsilon_{0}Q}{A}\left( \frac{t_{1}}{k_{1}} + \frac{t_{2}}{k_{2}} \right)$
C) $\frac{Q}{A\varepsilon_{0}}\left( \frac{k_{1}}{t_{1}} + \frac{k_{2}}{t_{2}} \right)$
D) $\frac{\varepsilon_{0}Q}{A}(k_{1}t_{1} + k_{2}t_{2})$
Explanation
(a) Potential difference across the condenser
$V = V_{1} + V_{2} = E_{1}t_{1} + E_{2}t_{2} = \frac{\sigma}{K_{1}\varepsilon_{0}}t_{1} + \frac{\sigma}{K_{2}\varepsilon_{0}}t_{2}$
$\Rightarrow V = \frac{\sigma}{\varepsilon_{0}}\left( \frac{t_{1}}{K_{1}} + \frac{t_{2}}{K_{2}} \right) = \frac{Q}{A\varepsilon_{0}}\left( \frac{t_{1}}{K_{1}} + \frac{t_{2}}{K_{2}} \right)$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.