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 ε_r = 4.

(i) Calculate capacitance of each capacitor if the equivalent capacitance of the combination is 4 μF.

(ii) Calculate the potential difference between the plates X and Y.

(iii) Estimate the ratio of electrostatic energies stored in X and Y.

Answer 1

(i) We know that the capacitance of a parallel-plate capacitor is given by

${\displaystyle C=\frac{{\varepsilon }_0{\varepsilon }_{r}A}{d}}$

${\displaystyle C_Y=\frac{{\varepsilon }_{0}4A}{d}}$

${\displaystyle C_X=\frac{{\varepsilon }_{0}A}{d}}$

Thus,

${\displaystyle \Rightarrow C_Y=4C_X}$

${\displaystyle \Rightarrow C_{eq}=\frac{C_x{C}_Y}{C_x+C_Y}=4\mu F}$

${\displaystyle \Rightarrow \frac{C_X\left(4C_X\right)}{5C_X}=4\mu F}$

${\displaystyle \Rightarrow C_X=5\mu F}$

${\displaystyle \Rightarrow C_Y=20\mu F}$

(ii) The potential difference between plates X and Y can be calculated as follows:

Q=CV

${\displaystyle \Rightarrow C_X{V}_X=C_Y{V}_Y}$

${\displaystyle \Rightarrow \frac{V_X}{V_Y}=\frac{C_X}{C_Y}=4}$

${\displaystyle \Rightarrow V_X=4V_Y}$

Also, V_Y+V_X=15

⇒V_Y=3 V

⇒V_X=12 V

(iii) The ratio of electrostatic energies can be calculated as follows:

${\displaystyle E=\frac{Q^2}{2C}}$

${\displaystyle \Rightarrow \frac{E_X}{E_Y}=\frac{C_Y}{C_X}=4}$

${\displaystyle \Rightarrow \frac{E_X}{E_Y}=\frac{4}{1}}$