Question

Two identical capacitors are connected in series as shown in the figure. A dielectric slab $(K>1)$ is placed between the plates of the capacitor $\text{B}$ and the battery remains connected. Which of the following statements(s) is/are correct following the insertion of the dielectric?

• A. The charge supplied by the battery increases
• B. The capacitance of the system increases
• C. The electric field in the capacitor $B$ increases
• D. The electrostatic potential energy decreases

Answer 1

Answer: (A), (B)

The capacitance of the system increases

On insertion of dielectric, capacitance increases $k$ times, thus, for system:

$C_{before}=\frac{C\times C}{C+C}=\frac{C}{2}$

$C_{after}=\frac{C\times \left(kC\right)}{C+\left(kC\right)}=\frac{kC}{k+1}$

$\because (k>1)$

$\therefore C_{after}>C_{before}$ $\left[\text{C increased}\right]$

$Q_{before}=\frac{CV}{2}or(V_B{)}_{before}=\frac{V}{2}$

$Q_{after}=\frac{kCV}{k+1}\text{or}(V_B{)}_{after}=\frac{V}{k+1}$

$\because (k>1)$

$\therefore (V_B{)}_{before}>(V_B{)}_{after}$ $\left[\text{V decreased}\right]$

Also, $E=\frac{V}{d}$

$\therefore (E_B{)}_{before}>(E_B{)}_{after}$ $\left[\text{V decreased}\right]$

Energy stored in the capacitor is,$U=\frac{1}{2}C{V}^2$

$\text{Thus}(U{)}_{after}>(U{)}_{before}$ $\left[\text{U increased}\right]$

Hence, option $A$ and $B$ is correct.