For the capacitor network shown in Fig. E24.28, the potential difference across ab is 48 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 differences across each capacitor

We know that the capacitance of a parallel plate capacitor is given by:

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

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

Step2: Calculating the equivalent capacitance

By solving the above equation, we get:

$\frac{1}{C_{eq}}=\frac{1}{C_1}+\frac{1}{C_2}$

Now, calculating the equivalent capacitance

$$\begin{gathered} C_{eq}=C \\ =\frac{C_1{C}_2}{C_1+C_2} \\ =\frac{\left(150\times {10}^{-9}F\right)\left(120\times {10}^{-9}F\right)}{150\times {10}^{-9}F+120\times {10}^{-9}F} \\ =66.6667\times {10}^{-9}F \end{gathered}$$

Now, using the equation for C we get:

$C=\frac{Q}{V_{ab}}$

Putting the values and solving for we get:

$$\begin{gathered} Q=C{V}_{ab} \\ =\left(66.667\times {10}^{-9}\right)\left(48\right) \\ =3.2\mu C \end{gathered}$$

Since the capacitors are connected in series, we the equivalent capacitance is$3.2\mu C$ for both

We that the potential energy stored in a capacitor U is equal to the work done while charging the capacitor:

$$\begin{gathered} U=W \\ =\frac{VQ}{2} \\ =\frac{Q^2}{2C} \end{gathered}$$

Substituting the values, we get:

$$\begin{gathered} U=\frac{{\left(3.2\times {10}^{-6}\right)}^2}{2\times 66.667\times {10}^{-9}} \\ =7.67\times {10}^{-5}J \end{gathered}$$

Therefore, the energy stored is$7.67\times {10}^{-5}J$

$$\begin{gathered} U_1=\frac{Q^2}{2C_1} \\ =\frac{{\left(3.2\times {10}^{-6}\right)}^2}{2\times 150\times {10}^{-9}} \\ =3.41\times {10}^{-5}J \end{gathered}$$

Similarly for the second capacitor $C_2$, stored energy is: $4.27\times {10}^{-5}J$

For potential difference we will use the formula:$V=\frac{Q}{C_1}$

Putting the values for both the capacitors we get:

$$\begin{gathered} V_1=21.33V \\ {V}_2=26.67V \end{gathered}$$

Therefore, the potential difference isrole="math" localid="1664252368798" $21.33V$ and$26.67V$