Question

A parallel-plate capacitor has plate area 25⋅0 cm² and a separation of 2⋅00 mm between the plates. The capacitor is connected to a battery of 12⋅0 V. (a) Find the charge on the capacitor. (b) The plate separation is decreased to 1⋅00 mm. Find the extra charge given by the battery to the positive plate.

Answer 1

Given:
Area of the plate, A = 25 cm² = 25 × 10^$-$4 m²
Separation between the plates, d = 2 mm = 2 × 10^$-$3 m
Potential difference between the plates, V = 12 V
The capacitance of the given capacitor is given by
$C=\frac{{\in }_{0}A}{d}$
$$\begin{gathered} =\frac{(8.85\times {10}^{-12})\times (25\times {10}^{-4})}{(2\times {10}^{-3})} \\ =11.06\times {10}^{-12}\mathrm{F} \end{gathered}$$

(a) Charge on the capacitor is given by
Q = CV
$$\begin{gathered} =11.06\times {10}^{-12}\times 12 \\ =1.33\times {10}^{-10}\mathrm{C} \end{gathered}$$

(b) When the separation between the plates is decreased to 1 mm, the capacitance C' can be calculated as:
$$\begin{gathered} C\text{'}=\frac{{\in }_0\mathrm{A}}{d} \\ =\frac{(8.85\times {10}^{-12})\times (25\times {10}^{-4})}{1\times {10}^{-3}} \\ =22.12\times {10}^{-12}\mathrm{F} \end{gathered}$$

Charge on the capacitor is given by
Q' = C'V
$$\begin{gathered} =22.12\times {10}^{-12}\times 12 \\ =2.65\times {10}^{-10}\mathrm{C} \\ \mathrm{Extra}\mathrm{charge} \\ =\left(2.65\times {10}^{-10}-1.32\times {10}^{-10}\right)\mathrm{C} \\ =1.33\times {10}^{-10}\mathrm{C} \end{gathered}$$