Question

A parallel-plate capacitor having plate area 25 cm² and separation 1⋅00 mm is connected to a battery of 6⋅0 V. Calculate the charge flown through the battery. How much work has been done by the battery during the process?

Answer 1

The capacitance of a parallel-plate capacitor is given by
$C=\frac{{\in }_{0}A}{d}$
Here,
A = Area of the plate
d = Distance between the parallel plates
Given:
A = 25 cm² = 25 × 10^$-$4 m²
d = 1.00 mm = 1 × 10^$-$3 m
Now,

$$\begin{gathered} C=\frac{{\in }_0\mathrm{A}}{d}=\frac{8.85\times {10}^{-12}\times 25\times {10}^{-4}}{1\times {10}^{-3}} \\ =2.21\times {10}^{-11}\mathrm{F} \end{gathered}$$

When the battery of voltage 6 V is connected to the capacitor, the charge (Q) that flows from the battery is equal to the amount of the charge that the given capacitor can hold.
$\Rightarrow$Q = CV
$$\begin{gathered} \Rightarrow Q=2.21\times {10}^{-11}\times 6.0 \\ =1.33\times {10}^{-10}\mathrm{C} \end{gathered}$$

The work done by the battery in charging the capacitor is calculated by taking the product of the magnitude of the charge transferred and the voltage of the battery.
Thus, we get
$$\begin{gathered} W=QV \\ =1.33\times {10}^{-10}\times 6.0 \\ =8.0\times {10}^{-10}\mathrm{J} \end{gathered}$$

Thus, the charge flown through the battery is 1.33 × 10^$-$10 C and the work done by the battery is 8.0 × 10^$-$10 J.