Question

Two capacitors of capacitance 20⋅0 pF and 50⋅0 pF are connected in series with a 6⋅00 V battery. Find (a) the potential difference across each capacitor and (b) the energy stored in each capacitor.

Answer 1

Given:

$$\begin{gathered} {\mathrm{C}}_1=20.0\mathrm{pF} \\ {\mathrm{C}}_2=50.0\mathrm{pF} \end{gathered}$$

When the capacitors are connected in series, their equivalent capacitance is given by
${\mathrm{C}}_{\mathrm{eq}}=\frac{C_1{C}_2}{C_1+C_2}$
∴ Equivalent capacitance, ${\mathrm{C}}_{\mathrm{eq}}=\frac{(50\times {10}^{-12})\times (20\times {10}^{-12})}{(50\times {10}^{-12})+(20\times {10}^{-12})}=1.428\times {10}^{-11}\mathrm{F}$

(a) The charge on both capacitors is equal as they are connected in series. It is given by

$$\begin{gathered} q=C_{\mathrm{eq}}\times V \\ \Rightarrow q=(1.428\times {10}^{-11})\times 6.0\mathrm{C} \\ \mathrm{Now}, \\ {V}_1=\frac{q}{{\mathrm{C}}_1}=\frac{(1.428\times {10}^{-11})\times 6.0\mathrm{C}}{(20\times {10}^{-12})} \\ \Rightarrow V_1=4.29\mathrm{V} \\ \mathrm{and} \\ {V}_2=\left(6.00-4.29\right)\mathrm{V}=1.71\mathrm{V} \end{gathered}$$

(b) The energies in the capacitors are given by
$$\begin{gathered} E_1=\frac{q^2}{2{\mathrm{C}}_1} \\ ={\left[(1.428\times {10}^{-11})\times 6.0\right]}^2\times \frac{1}{2\times 20\times {10}^{-12}} \\ =184\mathrm{pJ} \\ \mathrm{and} \\ {E}_2=\frac{q^2}{2{\mathrm{C}}_1} \\ ={\left[(1.428\times {10}^{-11})\times 6.0\right]}^2\times \frac{1}{2\times 50\times {10}^{-12}} \\ =73.5\mathrm{pJ} \end{gathered}$$