Question

The capacitance of a parallel plate capacitor is $C_a$ (fig). A dielectric of dielectric constant K is inserted as shown in fig. If $C_b$ and $C_c$ denote that capacitances in fig, then

• A. both $C_b,C_c>C_a$
• B. $C_c>C_a$ while $C_b>C_a$
• C. both $C_b,C_c<C_a$
• D. $C_a=C_b=C_c$

Answer 1

The correct option is A both $C_b,C_c>C_a$

For (a): the capacitance $C_a=\frac{A{ϵ}_0}{d}$

For (b): This will contain two capacitors and they are in series.

Here, $C_b=\frac{C_1{C}_2}{C_1+C_2}=\frac{\frac{A{ϵ}_0}{d/2}.\frac{AK{ϵ}_0}{d/2}}{\frac{A{ϵ}_0}{d/2}+\frac{AK{ϵ}_0}{d/2}}=\frac{2AK{ϵ}_0}{d(1+K)}=\frac{2C_a}{1+K}$

For (c): This will contain two capacitors and they are parallel.

Here, $C_c=C_1+C_2=\frac{\left(A/2\right){ϵ}_0}{d}+\frac{\left(A/2\right)K{ϵ}_0}{d}=\frac{\left(A/2\right){ϵ}_0}{d}[1+K]=\left(C_a/2\right)(1+K)$

As $K>1,C_b<C_a,C_c>C_a$