Question

Two parallel plate capacitors X and Y have the same area of plates and same separation between them. X has air between the plates while Y contains a dielectric medium of $E_r$ = 4.
(i) Calculate capacitance of each capacitor if equivalent capacitance of the combination is 4 $\mu F$.
(ii) Calculate the potential difference between the plates of
X and Y.
(iii) Estimate the ratio of electrostatic energy stored in X and Y.

Answer 1

Let the area of the plates be $A$ and distance between the plates be $d$

$\therefore$ Capacitance of capacitor X $C_1=\frac{A{ϵ}_o}{d}$

Also capacitance of capacitor Y $C_2=\frac{E_{r}A{ϵ}_o}{d}=\frac{4A{ϵ}_o}{d}=4C_1$

Equivalent capacitance in series combination, $C_{eq}=\frac{C_1\times C_2}{C_1+C_2}$

$\therefore$ $4\mu F=\frac{C_1\times 4C_1}{C_1+4C_1}$

OR $4\mu F=\frac{4C_1}{5}$ $⟹C_1=5\mu F$

$\therefore$ $C_2=4\times 5\mu F=20\mu F$

Let the total charge flowing through the circuit be $Q$

$\therefore$ $Q=C_{eq}V$

$\therefore$ $Q=4\mu F\times 15=60\mu C$

Potential drop across capacitor X, $V_1=\frac{Q}{C_1}=\frac{60\mu C}{5\mu F}=12V$

Potential drop across capacitor Y, $V_2=\frac{Q}{C_2}=\frac{60\mu C}{20\mu F}=3V$

Electrostatic energy stored in a capacitor $E=\frac{Q^2}{2C}$

$\therefore$ $\frac{E_1}{E_2}=\frac{\frac{Q^2}{2C_1}}{\frac{Q^2}{2C_2}}=\frac{C_2}{C_1}=\frac{20\mu F}{5\mu F}=4$