Question

Each of the plates shown in figure (31-E19) has surface area $\left(96/{\epsilon }_0\right)\times {10}^{-12}$ Fm on one side and the separation between the consecutive plates is 4⋅0 mm. The emf of the battery connected is 10 volts. Find the magnitude of the charge supplied by the battery to each of the plates connected to it.

Answer 1

Here, three capacitors are formed and each of them has area A.
Now,
$$\begin{gathered} A=\frac{96}{{\in }_0}\times {10}^{-12}\mathrm{F}-\mathrm{m} \\ d=4\mathrm{mm} \\ =3\times {10}^{-3}\mathrm{m} \end{gathered}$$

The capacitance of the capacitors is given by
$C=\frac{{\in }_{0}A}{d}=\frac{{\in }_0\left(96/{\in }_0\right)\times {10}^{-12}}{4\times {10}^{-3}}=24\times {10}^{-9}\mathrm{F}$

Because the three capacitors are arranged in series, the equivalent capacitance is given by

$C_{\mathrm{eq}}=\frac{24\times {10}^{-9}}{3}=8\times {10}^{-9}\mathrm{F}$

The total charge on the capacitor is given by
Q = C_eqV
$$\begin{gathered} \Rightarrow Q=8\times {10}^{-9}\times 10 \\ \Rightarrow Q=8\times {10}^{-8}C \end{gathered}$$

The charge on a single plate is given by
$$\begin{gathered} Q_{\mathrm{P}}=2\times 8\times {10}^{-8} \\ \Rightarrow Q_{\mathrm{P}}=16\times {10}^{-8}=0.16\mathrm{\mu C} \end{gathered}$$