Electrostatics — Easy JEE physics MCQ
A parallel plate capacitor with plate area \(A\) and separation \(d\) is charged to a potential difference \(V\). A dielectric slab of dielectric constant \(K\) and thickness \(d\) is inserted between the plates while the capacitor remains connected to the battery. The ratio of the final energy stored to the initial energy stored is:
- A. \(\frac{1}{K}\)
- B. \(K\)
- C. \(K^2\)
- D. \(\frac{1}{K^2}\)
Solution
Initial capacitance: \(C_0 = \frac{\varepsilon_0 A}{d}\)
Final capacitance with dielectric: \(C = KC_0\)
Since the capacitor remains connected to the battery, voltage \(V\) remains constant.
Initial energy: \(U_i = \frac{1}{2}C_0 V^2\)
Final energy: \(U_f = \frac{1}{2}CV^2 = \frac{1}{2}KC_0 V^2\)
Ratio:
\[
\frac{U_f}{U_i} = \frac{\frac{1}{2}KC_0 V^2}{\frac{1}{2}C_0 V^2} = K
\]
So the correct option is B.