Question

In Fig. 25-36, the capacitances are${\mathbf{C}}_{\mathbf{1}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{\mu F}$and ${\mathbf{C}}_{\mathbf{2}}\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{\mu F}$, and both capacitors are charged to a potential difference of$\mathbf{V}\mathbf{=}\mathbf{100}\mathbf{}\mathbf{V}$but with opposite polarity as shown. Switches S1 and S2 are now closed. (a) What is now the potential difference between points aand b? What now is the charge on (b) capacitor 1 and (c) capacitor 2?

Short answer

1. The potential difference between points a and b is ${\mathrm{V}}_{\mathrm{ab}}=50\mathrm{V}$
2. The charge on the capacitor 1 is ${\mathrm{q}}_1=50\mathrm{\mu C}$
3. The charge on the capacitor 2 is ${\mathrm{q}}_2=150\mathrm{\mu C}$

Step-by-step solution

Step 1: Given

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

Determining the concept

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,${\mathrm{C}}_{\mathrm{eq}}=\sum _{\mathrm{j}=1}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{j}}$

For series combination,

$\frac{1}{{\mathrm{C}}_{\mathrm{eq}}}=\sum _{\mathrm{j}=1}^{\mathrm{n}}\frac{1}{{\mathrm{C}}_{\mathrm{j}}}$

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

(a) Determining the potential difference between points a  and  b

It is given that,

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

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

$$\begin{gathered} {\mathrm{q}}_1={\mathrm{C}}_1\mathrm{V} \\ =1.0\mathrm{\mu F}\times 100\mathrm{V} \\ =100\mathrm{\mu C} \end{gathered}$$

Similarly, the charge on the capacitor 2 is,

$$\begin{gathered} {\mathrm{q}}_2={\mathrm{C}}_2\mathrm{V} \\ =3.0\mathrm{\mu F}\times 100\mathrm{V} \\ =300\mathrm{\mu C} \end{gathered}$$

So the net charge on the combination is,

$$\begin{gathered} \mathrm{q}={\mathrm{q}}_2-{\mathrm{q}}_1 \\ =300\mathrm{\mu C}-100\mathrm{\mu C} \\ =200\mathrm{\mu C} \end{gathered}$$

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

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

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

$$\begin{gathered} {\mathrm{V}}_{\mathrm{ab}}=\frac{\mathrm{q}}{{\mathrm{C}}_{\mathrm{eq}}} \\ =\frac{200\mathrm{\mu C}}{4\mathrm{\mu F}} \\ =50\mathrm{V} \end{gathered}$$

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

(b) Determining the charge of the capacitor  1

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

$$\begin{gathered} {\mathrm{q}}_1={\mathrm{C}}_1{\mathrm{V}}_{\mathrm{ab}} \\ =1.0\mathrm{\mu F}\times 50\mathrm{V} \\ =50\mathrm{\mu C} \end{gathered}$$

Hence, the charge on the capacitor 1 is $q_1=50\mathrm{\mu C}$.

(c) Determining the charge of the capacitor  

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

$$\begin{gathered} {\mathrm{q}}_2={\mathrm{C}}_2{\mathrm{V}}_{\mathrm{ab}} \\ =3.0\mathrm{\mu F}\times 50\mathrm{V} \\ =150\mathrm{\mu C} \end{gathered}$$

Hence, the charge on the capacitor 2 is ${\mathrm{q}}_2=150\mathrm{\mu C}$.

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.