Question

In the figure shown capacitors A and B of capacitance C each are in steady state. A dielectric slab of dielectric constant K = 2 and dimensions equal to the inner dimensions of the capacitor is inserted in the space between the plates of the capacitor B. Choose the correct option(s):

• A. Charge on each capacitor will increase by $\frac{CE}{6}$.
  
• B. In the process of inserting the dielectric, energy of the cell decreases by an amount of $\frac{C{E}^2}{6}$.
• C. In the process of inserting the dielectric, energy of the cell increases by an amount of $\frac{C{E}^2}{6}$
  
• D. In the process of inserting the dielectric, the energy in the capacitor A increases by amount of $\frac{7C{E}^2}{72}$..

Answer 1

Answer: (A), (B), (D)

The correct options are
A

Charge on each capacitor will increase by $\frac{CE}{6}$.

B

In the process of inserting the dielectric, energy of the cell decreases by an amount of $\frac{C{E}^2}{6}$.

D

In the process of inserting the dielectric, the energy in the capacitor A increases by amount of $\frac{7C{E}^2}{72}$..

Initially, $Q_A=Q_B=\frac{CE}{2},$ when dielectric is inserted.
$Q_A^{\prime }=Q_B^{\prime }=\frac{2CE}{3},\therefore △Q=\frac{2CE}{3}-\frac{CE}{2}=\frac{CE}{6}$
Work done by battery=$\frac{CE}{6}\times E=\frac{C{E}^2}{6}△U_A-U_i=\frac{Q_A^{\prime 2}}{2C}-\frac{Q_A^2}{2C}=\frac{7C{E}^2}{72}$