Question

A parallel-plate capacitor with no dielectric has a capacitance of $0.5\mu F$ . The space between the plates is then filled with equal amounts of two dielectrics of dielectric constants $2$ and $3$ in the two arrangements as shown one by one. Find the ratio of capacitances $C_I$ and $C_{II}$

Answer 1

The first case can be considered as the capacitors in parallel

Equivalent capacitance is equal to

$C_A=k_1{C}_1+k_2{C}_2$

$=2\left(\frac{A{\epsilon }_0}{2d}\right)+3\left(\frac{A{\epsilon }_0}{2d}\right)$

$=\frac{5}{2}\times 0.5\mu F$

The second case can be considered as the capacitors in series

Equivalent capacitance is equal to

$\frac{1}{C}=\frac{1}{k_1{C}_1}+\frac{1}{k_2{C}_2}$

$=\frac{d}{2\times 2A{\epsilon }_0}+\frac{d}{3\times 2A{\epsilon }_0}$

$C_B=\frac{12}{5}\times 0.5\mu F$

$\frac{C_A}{C_B}=\frac{25}{24}$