Question

In Fig. 25-30, the battery has a potential difference of V = 10.0 V, and the five capacitors each have a capacitance of$\mathbf{10}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{\mu F}$What is the charge on (a) capacitor 1 and (b) capacitor 2?

Short answer

a) The charge on the capacitor $1\mathrm{is}{\mathrm{q}}_1={10}^{-4}\mathrm{C}$

b) The charge on the capacitor $2\mathrm{is}{\mathrm{q}}_2=2\times {10}^{-5}\mathrm{C}$

Step-by-step solution

Step 1: Given data

The potential difference is V = 10 V

${\mathrm{C}}_1={\mathrm{C}}_2={\mathrm{C}}_3={\mathrm{C}}_4={\mathrm{C}}_5=10\mathrm{\mu F}\left(\frac{{10}^{-6}\mathrm{F}}{1\mathrm{\mu F}}\right)=1.0\times {10}^{-5}\mathrm{F}$

Determining the concept

Find the charge on capacitor1by using the concept of capacitance. To find the charge on capacitor2, find the equivalent capacitance and potential difference across capacitors2.

Formulae are as follows:

q = CV

For parallel combination,${\mathbf{C}}_{\mathbf{eq}}\mathbf{=}\mathbf{\sum }_{\mathbf{j}\mathbf{=}\mathbf{1}}^{\mathbf{n}}{\mathbf{C}}_{\mathbf{j}}$

For series combination,$\frac{\mathbf{1}}{{\mathbf{C}}_{\mathbf{eq}}}\mathbf{=}\mathbf{\sum }_{\mathbf{j}\mathbf{=}\mathbf{1}}^{\mathbf{n}}\frac{\mathbf{1}}{{\mathbf{C}}_{\mathbf{j}}}$

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

(a) Determining the charge of the capacitor

It is known that,

q = CV

The potential difference across the capacitor ${\mathrm{C}}_1$is,

${\mathrm{V}}_1=10\mathrm{V}$

So the charge on the capacitor 1 is,

$$\begin{gathered} {\mathrm{q}}_1={\mathrm{C}}_1{\mathrm{V}}_1 \\ {\mathrm{q}}_1=1.0\times {10}^{-5}\mathrm{F}\times 10\mathrm{V} \\ =1.0\times {10}^{-4}\mathrm{C} \end{gathered}$$

Hence, the charge on the capacitor 1 is $1.0\times {10}^{-4}\mathrm{C}$

(b) Determining the charge of the capacitor

For finding the charge${\mathrm{q}}_2$, first, find the equivalent capacitance.

Consider the three-capacitor combination consisting of${\mathrm{C}}_2$and its two closest neighbors, each of capacitance C. Using the formula for parallel and series combination of the capacitor, write the equivalent capacitance of this combination as

${\mathrm{C}}_{\mathrm{eq}}=\mathrm{C}+\frac{{\mathrm{C}}_2\mathrm{C}}{\mathrm{C}+{\mathrm{C}}_2}$

By substituting the values,

$$\begin{gathered} {\mathrm{C}}_{\mathrm{eq}}=\mathrm{C}+\frac{{\mathrm{C}}_{\times }\mathrm{C}}{\mathrm{C}+\mathrm{C}} \\ {\mathrm{C}}_{\mathrm{eq}}=\mathrm{C}+\frac{{\mathrm{C}}^2}{2\mathrm{C}} \\ =\frac{3{\mathrm{C}}^2}{2\mathrm{C}} \\ =1.5\mathrm{C} \end{gathered}$$

The voltage drop in this combination is,

$\mathrm{V}=\frac{\mathrm{CV}}{\mathrm{C}+{\mathrm{C}}_{\mathrm{eq}}}$

By outing the value of ${\mathrm{C}}_{\mathrm{eq}}$,

$$\begin{gathered} \mathrm{V}=\frac{\mathrm{CV}}{\mathrm{C}+1.5\mathrm{C}} \\ =\frac{{\mathrm{CV}}_1}{2.5\mathrm{C}} \\ =0.4{\mathrm{V}}_1 \end{gathered}$$

This voltage difference is divided into two equal parts between ${\mathrm{C}}_2$and the capacitor connected in series with it. So, the total voltage across the capacitor 2 is,

$$\begin{gathered} {\mathrm{V}}_2=\frac{\mathrm{V}}{2} \\ =\frac{0.4{\mathrm{V}}_1}{2} \\ =0.2{\mathrm{V}}_1 \end{gathered}$$

Thus, the total charge on the capacitor 2 will be,

$$\begin{gathered} {\mathrm{q}}_2={\mathrm{C}}_2{\mathrm{V}}_2 \\ {\mathrm{q}}_2=1.0\times {10}^{-6}\mathrm{F}\times 0.2{\mathrm{V}}_1 \end{gathered}$$

By substituting the value of${\mathrm{V}}_1$,

$$\begin{gathered} {\mathrm{q}}_2=1.0\times {10}^{-5}\mathrm{F}\times 0.2\times 10\mathrm{V} \\ =2\times {10}^{-5}\mathrm{C} \end{gathered}$$

Hence, the charge on the capacitor 2 is$2\times {10}^{-5}\mathrm{C}$

Therefore, we can find the charge on both capacitors by using the concept of capacitance.