91Ó°ÊÓ

$$\begin{gathered} {\mathrm{C}}_1=1.0\mathrm{μ¹ó} \\ {\mathrm{C}}_2=3.0\mathrm{μ¹ó} \\ \mathrm{V}=100\mathrm{V} \end{gathered}$$

After the switches are closed, the potential differences across the capacitors are the same, and they are connected in parallel. To find the equivalent capacitance. Find the potential difference and charge on each capacitor by using the relation between capacitance and charge.

Formulae are as follows:

$\mathrm{q}=\mathrm{CV}$

For parallel combination,${\mathrm{C}}_{\mathrm{eq}}=\underset{\mathrm{j}=1}{\overset{\mathrm{n}}{∑}}{\mathrm{C}}_{\mathrm{j}}$

For series combination,

$\frac{1}{{\mathrm{C}}_{\mathrm{eq}}}=\underset{\mathrm{j}=1}{\overset{\mathrm{n}}{∑}}\frac{1}{{\mathrm{C}}_{\mathrm{j}}}$

Where C is capacitance, V is the potential difference, and q is the charge.

It is given that,

$\mathrm{q}=\mathrm{CV}$

We can calculate the charge on the capacitor 1 due to potential$\mathrm{V}=100\mathrm{V}$

$$\begin{gathered} {\mathrm{q}}_1={\mathrm{C}}_1\mathrm{V} \\ =1.0\mathrm{μ¹ó}×100\mathrm{V} \\ =100\mathrm{μ°ä} \end{gathered}$$

Similarly, the charge on the capacitor 2 is,

$$\begin{gathered} {\mathrm{q}}_2={\mathrm{C}}_2\mathrm{V} \\ =3.0\mathrm{μ¹ó}×100\mathrm{V} \\ =300\mathrm{μ°ä} \end{gathered}$$

So the net charge on the combination is,

$$\begin{gathered} \mathrm{q}={\mathrm{q}}_2-{\mathrm{q}}_1 \\ =300\mathrm{μ°ä}-100\mathrm{μ°ä} \\ =200\mathrm{μ°ä} \end{gathered}$$

Since, both the capacitance are in parallel, find its equivalent capacitance.

$$\begin{gathered} {\mathrm{C}}_{\mathrm{eq}}={\mathrm{C}}_1+{\mathrm{C}}_2 \\ =1.0\mathrm{μ¹ó}+3.0\mathrm{μ¹ó} \\ =4.0\mathrm{μ¹ó} \end{gathered}$$

Now, the potential difference across points a and b is,

$$\begin{gathered} {\mathrm{V}}_{\mathrm{ab}}=\frac{\mathrm{q}}{{\mathrm{C}}_{\mathrm{eq}}} \\ =\frac{200\mathrm{μ°ä}}{4\mathrm{μ¹ó}} \\ =50\mathrm{V} \end{gathered}$$

Hence, the potential difference between points a and b is ${\mathrm{V}}_{\mathrm{ab}}=50\mathrm{V}$.

The charge on the capacitor 1 due to potential ${\mathrm{V}}_{\mathrm{ab}}=50\mathrm{V}$is,

$$\begin{gathered} {\mathrm{q}}_1={\mathrm{C}}_1{\mathrm{V}}_{\mathrm{ab}} \\ =1.0\mathrm{μ¹ó}×50\mathrm{V} \\ =50\mathrm{μ°ä} \end{gathered}$$

Hence, the charge on the capacitor 1 is $q_1=50\mathrm{μ°ä}$.

The charge on the capacitor 2 due to potential ${\mathrm{V}}_{\mathrm{ab}}=50\mathrm{V}$is,

$$\begin{gathered} {\mathrm{q}}_2={\mathrm{C}}_2{\mathrm{V}}_{\mathrm{ab}} \\ =3.0\mathrm{μ¹ó}×50\mathrm{V} \\ =150\mathrm{μ°ä} \end{gathered}$$

Hence, the charge on the capacitor 2 is ${\mathrm{q}}_2=150\mathrm{μ°ä}$.

Therefore, by using the relation between capacitance and charge and the concept of equivalent capacitance find the potential difference between points a and b and charges on both the capacitors.