Solution
Let the total area be $A$. The dielectric covers $A/3$ (capacitance $C_1$) and air covers $2A/3$ (capacitance $C_2$). The total charge $Q$ is constant (battery disconnected). The two sections are in parallel, so potential $V’$ is the same.
1. Dielectric Constant (Statement a):
Charges are equal: $q_1 = q_2$.
Since $q = CV$, and $V$ is common: $C_1 = C_2$.
$$\frac{k\epsilon_0 (A/3)}{d} = \frac{\epsilon_0 (2A/3)}{d}$$
$$\frac{k}{3} = \frac{2}{3} \implies k = 2.0$$
Statement (a) is Correct.
2. Bound Charge (Statement b):
The bound charge density is $\sigma_p = \sigma_{free} (1 – 1/k)$.
Charge $q_p = q_1 (1 – 1/k)$.
Given $q_1 = Q/2 = (C_0 V)/2$. And $k=2$.
$$q_p = \frac{C_0 V}{2} \left(1 – \frac{1}{2}\right) = \frac{C_0 V}{4} = 0.25 C_0 V$$
Statement (b) is Correct.
3. Electrostatic Force (Statements c & d):
The force on a conducting plate element with surface charge density $\sigma_{free}$ is determined by the field in the vacuum gap just outside the metal surface: $F = \frac{\sigma_{free}^2}{2\epsilon_0} \times \text{Area}$.
- Force on Dielectric Part ($F_1$): $$F_1 = \frac{(q_1 / (A/3))^2}{2\epsilon_0} \cdot \frac{A}{3} = \frac{9 q_1^2}{2\epsilon_0 A} \cdot \frac{1}{3} = \frac{3 (Q/2)^2}{2\epsilon_0 A} = \frac{3 Q^2}{8 \epsilon_0 A}$$
- Force on Air Part ($F_2$): $$F_2 = \frac{(q_2 / (2A/3))^2}{2\epsilon_0} \cdot \frac{2A}{3} = \frac{9 q_2^2}{8\epsilon_0 A} \cdot \frac{2}{3} = \frac{3 (Q/2)^2}{4\epsilon_0 A} = \frac{3 Q^2}{16 \epsilon_0 A}$$
Clearly $F_1 \neq F_2$ (Statement c is incorrect).
Total Force Ratio: $$F_{new} = F_1 + F_2 = \frac{6 Q^2}{16 \epsilon_0 A} + \frac{3 Q^2}{16 \epsilon_0 A} = \frac{9 Q^2}{16 \epsilon_0 A}$$ Original Force (no dielectric): $$F_{old} = \frac{Q^2}{2\epsilon_0 A} = \frac{8 Q^2}{16 \epsilon_0 A}$$ Ratio: $$\frac{F_{new}}{F_{old}} = \frac{9}{8}$$ Statement (d) is Correct.
Answer: (a), (b) and (d)