Question

The Figure shows capacitors $A$ and $B$ of capacitance $C$ each. A dielectric slab of dielectric constant $k=2$ and same dimensions as the capacitors is inserted between the plates of the capacitor $B$. In the process of inserting the dielectric slab:

• A. Work done by battery is $\frac{c{ϵ}^2}{b}$
• B. Energy of capacitor B increases by $\frac{c{ϵ}^2}{72}$
• C. Energy of capacitor A increases by $\frac{7c{ϵ}^2}{72}$
• D. Energy of cell increases

Answer 1

Answer: (A), (C)

The correct options are
A Work done by battery is $\frac{c{ϵ}^2}{b}$
C Energy of capacitor A increases by $\frac{7c{ϵ}^2}{72}$

Charge on the capacitors without dielectric $q_i=\frac{Cϵ}{2}$
Charge on the capacitors with dielectric $q_f=\frac{2Cϵ}{3}$
Work done by the battery $\text{W}=ϵ(q_f-q_i)=\frac{C{ϵ}^2}{6}$
Initial energy of A $\Rightarrow U_i=\frac{C{ϵ}^2}{8}$
Final energy of A $\Rightarrow U_f=\frac{2C{ϵ}^2}{9}$
Initial energy of B $\Rightarrow U_i=\frac{C{ϵ}^2}{8}$
Final energy of B $\Rightarrow U_f=\frac{C{ϵ}^2}{9}$