Question

In the figure shown $C_1=4\mu F$ (without dielectric) and $C_2=4\mu F$ (with a dielectric slab of dielectric constant $K=2$). Now the same slab after removing from $C_2$ is filled in $C_1$. Then match the following two columns:

Answer 1

After removing slam from $C_2$ and put into $C_1$, $C_1=2\times 4=8\mu F$ and $C_2=\frac{4}{2}=2\mu F$.

A) initial the equivalent capacitance $C_i=\frac{4\times 4}{4+4}=2\mu F$ and $Q_i=C_{i}V$

and final equivalent capacitance $C_f=\frac{8\times 2}{8+2}=1.6\mu F$ and $Q_f=C_{f}V$

As the capacitors are in series so charge on each capacitor is equal to the equivalent charge. As equivalent capacitance is decreasing so charge on both capacitors will decrease.

B) Initial energy in $C_2$ is $U_{2i}=\frac{1}{2}\frac{Q_i^2}{C_2}=\frac{1}{2}\times \frac{4V^2}{4}=\frac{V^2}{2}$

Final energy $U_{2f}=\frac{1}{2}\frac{Q_f^2}{C_2}=\frac{1}{2}\times \frac{(1.6V{)}^2}{4}=\frac{V^2}{2}\left(1.28\right)$

since $U_{2f}>U_{2i},$ energy will increase.

C) $V_{2i}=\frac{Q_i}{C_2}=\frac{2V}{4}=V/2$ and $V_{2f}=\frac{Q_f}{C_2}=\frac{1.6V}{2}$

since $V_{2f}>V_{2i},$ potential across $C_2$ will increase.

D) $E_{2f}=\frac{V_{2f}}{d}$ , As d is constant here so $E_{2f}\propto V_{2f}.$ Thus, electric will also increase.