Question

The plates of a parallel plate capacitor with no dielectric are connected to a voltage source. Now a dielectric of dielectric constant $K$ is inserted to fill the whole space between the plates with voltage source remaining connected to the capacitor.
Choose the correct statements.

• A. The energy stored in the capacitor will become $K-$ times
• B. The electric field inside the capacitor will decrease $K-$ times
• C. The force of attraction between the plates will become $K^2-$ times
• D. The charge on the capacitor will become $K-$ times

Answer 1

Answer: (A), (C), (D)

The correct options are
A The energy stored in the capacitor will become $K-$ times
C The force of attraction between the plates will become $K^2-$ times
D The charge on the capacitor will become $K-$ times

Let $Q_i,U_i,E_i,F_i$ denote charge,energy stored,Electric field, force between plates of a capacitor without dielectric respectively and
$Q_f,U_f,E_f,F_f$ denote charge,energy stored,Electric field, force between plates of a capacitor with dielectric (constacnt $K$) respectively.

Since the capacitor is connected to the battery, the voltage across the capacitor remains constant.
Energy stored:
$U_i=\frac{1}{2}C{V}^2$
$U_f=\frac{1}{2}\left(KC\right)V^2$ (Capacitance increases by $K$ due to the presence of dielectric)
$U_f=K{U}_i$ ($A$ is correct)
Charge:
$Q_i=CV$
$Q_f=KCV$
$Q_f=K{Q}_i$ ($D$ is correct)
Electric field between the plates:
$E_i=\frac{Q_i}{{ϵ}_{0}A}$
$E_f=\frac{Q_f}{{ϵ}_{0}A}$
$E_f=\frac{K{Q}_i}{{ϵ}_{0}A}$
$E_f=K{E}_i$ ($B$ is wrong)
Force between the plates:
$F_i=Q_i\left(\frac{E_i}{2}\right)$
(Here Electric is field due to one of the plates)
$F_f=Q_f\left(\frac{E_f}{2}\right)$
$F_f=K{Q}_i\left(\frac{K{E}_i}{2}\right)$
$F_f=K^2{F}_i$ ($C$ is correct).