Question

An isolated capacitor has charge \(Q\) and plate area \(A\). 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

Answer 1

Diagram
Potential difference between the plates is
\(V=E_1 t_1+E_2 t_2\)
(\(E\) = electric field between plates )
\(=\dfrac{Qt_1}{k_1 A\epsilon_0~ }+\dfrac{Qt_2}{k_2 A\epsilon_0 }\)
Thus, net potential difference across condenser is
= \(\dfrac{Q}{A\epsilon_0}\left ( \dfrac{t_1}{k_1}+\dfrac{t_2}{k_2} \right )\)
Hence, correct option is (A)