Question

In the figure shown, capacitors $A$ and $B$ of capacitance $C$ 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 $B$. In steady state, choose the correct options

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


Increase in charge on each capacitor $=\frac{2CE}{3}-\frac{CE}{2}=\frac{CE}{6}$

Work done by the battery in the battery = charge flowing through the battery $\times$ emf $=\frac{CE}{6}\times E=\frac{C{E}^2}{6}$

Increase in energy stored in capacitor $A=\frac{1}{2}C(\frac{2E}{3}{)}^2-\frac{1}{2}C(\frac{E}{2}{)}^2=\frac{7C{E}^2}{72}$