Question

For the capacitor network shown in Fig. E24.29, the potential difference across ab is 220 V. Find (a) the total charge stored in this network; (b) the charge on each capacitor; (c) the total energy stored in the network; (d) the energy stored in each capacitor; (e) the potential difference across each capacitor.

Short answer

(a) For the capacitor the potential difference across ab of $220V$, the total charge stored in this network is $24.2\mu C$

(b) The charge on each capacitor is $7.70\mu C$and $16.5\mu C$

(c) The total energy stored in the network is $2.66mJ$

(d) The potential difference across the capacitor is $220.0V$

Step-by-step solution

Calculating total charge on the capacitor

Capacitance of a parallel plate capacitor is given by:

$\begin{array}{r}C=\frac{Q}{V_{ab}}\\ ={\epsilon }_0\frac{A}{d}\end{array}$

Where, $C$is the capacitance, $V_{ab}$is the potential difference between the plates, $Q$is the magnitude of the charge, $d$is the distance between the plates, $A$is the area of each plates and ${\epsilon }_0$is the electric constant whose value is$8.85x{10}^{12}F/m$

Now we know that the equivalent capacitance of capacitor in parallel is given by:

$\begin{array}{r}{C}_{\text{eq}}=C_1+C_2\\ =35+75\\ =110\times {10}^{-9}\mathrm{F}\end{array}$

The charge is given by:

$\begin{array}{r}C=\frac{Q}{V_{ab}}\\ Q_{\text{tobla}}=C{V}_{ab}\\ =\left(110\times {10}^{-9}\mathrm{F}\right)\left(220\mathrm{V}\right)\\ =2.24\times {10}^{-5}\mathrm{C}\end{array}$

Therefore, the total charge is $2.42x{10}^{-5}C$

Potential difference

Now it is given that

$\begin{array}{r}{C}_1=35\times {10}^{-9}\mathrm{F}\\ C_2=75\times {10}^{-9}\mathrm{F}\end{array}$

Since, $V=220V$is same for the both capacitors, as both are connected in parallel.

Hence,

localid="1668315618253" $\begin{array}{r}C=Q/V_{ab}\\ Q=CV\end{array}$

Now,$\begin{array}{r}{Q}_1=C_1{V}_1\\ =C_{1}V\\ =\left(35\times {10}^{-9}\mathrm{F}\right)\left(220\mathrm{V}\right)\\ =7.7\times {10}^{-6}\mathrm{C}\end{array}$

Similarly, $Q_2=16.5x{10}^{-6}C$

Potential energy stored

We know that the potential energy stored in a capacitor is given by:

$\begin{array}{r}U=W\\ =\frac{VQ}{2}\\ =\frac{Q^2}{2C}\\ =\frac{1}{2}C{V}^2\end{array}$

Where, $U$is the potential energy stored, $W$is the work done,$Q$is the charge, $C$is capacitance and V is the potential difference between the plates.

Now,

$\begin{array}{r}U=\frac{1}{2}C{V}^2\\ U_{\text{tobis}}=\frac{1}{2}{C}_{\text{eq}}{V}_{\text{total}}^2\\ =\frac{1}{2}\left(110\times {10}^{-19}\mathrm{F}\right)(220\mathrm{V}{)}^2\\ =2.662\times {10}^{-3}\mathrm{J}\end{array}$

Now, energy stored in each capacitor is:

$\begin{array}{r}{U}_1=\frac{1}{2}{C}_1{V}_1^2\\ =\frac{1}{2}\left(35\times {10}^{-9}\right)(220\mathrm{V}{)}^2\\ =8.47\times {10}^{-4}\mathrm{J}\end{array}$

Similarly for the other capacitor is:

$\begin{array}{r}{U}_2=\frac{1}{2}{C}_2{V}_2^2\\ =\frac{1}{2}\left(75\times {10}^{-9}\right)(220\mathrm{V}{)}^2\\ =18.15\times {10}^{-4}\mathrm{J}\end{array}$

Therefore, we can conclude that the potential difference is same for both the capacitor.