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 $B$ and the battery remains connected. Which of the following statement(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 correct options are
A The charge supplied by the battery increases.
B The capacitance of the system increases.

Equivalent capacitance Before insertion of di-electric :-
$C_{eq}=\frac{c_1{c}_2}{c_1+c_2}=\frac{C\times C}{C+C}=\frac{C}{2}$
Equivalent capacitance after insertion of di-electric :-
On insertion of dielectric, capacitance increased $k$ times, Thus, for system :-
$C_{\text{After}}=\frac{C\times \left(KC\right)}{C+KC}=\frac{KC}{K+1}$
[As $K>1$]
As Battery remains connected, potential - difference $\left[V\right]$ remains constant
$Q=CV$
$\therefore C_{\text{After}}>C_{\text{Before}}$ [$C$ increased]
$\therefore Q_{\text{After}}>Q_{\text{Before}}$ [$Q$ increased]
Potential difference and charge on capacitor before di-electric on $B$ :
$Q_{\text{Before}}=\frac{CV}{2}$
$V_B=\frac{V}{2}$
Potential difference and charge on capacitor after insertion of di-electric :
$Q_{\text{After}}=C_{eq}V=\frac{KCV}{K+1}$
$V_{B_{\text{After}}}=\frac{Q_{\text{After}}}{KC}=\frac{\left[\frac{KCV}{K+1}\right]}{KC}$
$V_{B_{\text{After}}}=\left[\frac{V}{K+1}\right][K>1]$
[Hence, $V$ decreases]
$[V_B{]}_{\text{Before}}>[V_B{]}_{\text{After}}$
and $E=\frac{V}{d}$
hence, $[E_B{]}_{\text{Before}}>[E_B{]}_{\text{After}}$ [$E$ decreases]
Energy stored in a capacitor:-
$U=\frac{1}{2}C{V}^2$
As $V=$ constant
$C_{\text{After}}>C_{\text{Before}}$.
Hence $U_{\text{After}}>U_{\text{Before}}$
So, Electrostatic Potential energy increases.
Hence; option (A) and (B) are correct.