Question

Two capacitors $C_1=8\mu F$ and $C_2=4\mu F$ are connected in series between points A and B. An ideal cell of emf $12\text{V}$ is connected between A and B for a long time. Now, a slab of dielectric constant $k=2$ is slowly introduced, fully in the gap of capacitor $C_2$.

• A. As the dielectric is introduced, the energy taken from the cell is $192\mu J$.
• B. As the dielectric is introduced the increase in electrostatic energy stored in the capacitors is $96\mu J$.
• C. The work done by dielectric slab on the filling agent during the filling process is positive.
• D. The work done by dielectric slab on the filling agent during the filling process is zero.

Answer 1

Answer: (A), (B), (C)

The work done by dielectric slab on the filling agent during the filling process is positive.

Before introducing the dielectric, the equivalent capacitance of the circuit is $C_i=\frac{C_1{C}_2}{C_1+C_2}=\frac{32}{12}=\frac{8}{3}\mu F.$
After introducing the dielectric, the equivalent capacitance of the circuit is
$C_f=\frac{k{C}_1{C}_2}{C_1+k{C}_2}=\frac{64}{16}=4\mu F$.
Change in charge flowing through battery
$\Delta q=\Delta CV=(4-\frac{8}{3})12=16\mu C$
Energy taken from the cell = charge flown through the cell x cell voltage $=16\times 12=192\mu J.$

Electrostatic energy stored in capacitors initially is $U_i=\frac{1}{2}C{V}^2=\left(\frac{1}{2}\right)\times \frac{8}{3}\times {12}^2=192\mu J$.
Electrostatic energy stored in capacitors after introducing dielectrc is
$U_f=\frac{1}{2}C{V}^2=\left(\frac{1}{2}\right)\times 4\times {12}^2=288\mu J$.
Therefore, increase in the electrostatic energy stored in the capacitors $=96\mu J$.