Question

Plot 1 in Fig. 25-32agives the charge qthat can be stored on capacitor 1 versus the electric potential Vset up across it. The vertical scale is set by${\mathit{q}}_{\mathbf{s}}\mathbf{=}\mathbf{16}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{\mu C}$and the horizontal scale is set by${\mathbf{\text{V}}}_{\mathbf{s}}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{V}$Plots 2 and 3 are similar plots for capacitors 2 and 3, respectively. Figure bshows a circuit with those three capacitors and abattery. What is the charge stored on capacitor 2 in that circuit?

Short answer

The charge stored on the capacitor 2 is ${\mathrm{q}}_2=12\mathrm{\mu C}$.

Step-by-step solution

Step 1: Given data

The vertical scale set is by${\mathrm{q}}_{\mathrm{s}}=16\mathrm{\mu C}$

The horizontal scale is set by${\mathrm{V}}_{\mathrm{s}}=2\mathrm{V}$

The voltage across the battery is V = 6 V

Determining the concept

Capacitance is the ratio of charge to potential. So the slope of the charge versus the potential graph gives us the capacitance for each capacitor. Find the equivalent capacitance by using the formulae for series and parallel combinations. By using the formula for capacitance, find the charge stored by capacitor 2.

Formulae are as follows:

q = 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, q is the charge on the particle

Determining the charge stored on the capacitor

From the graph, find the slope of each line which will give us the capacitance for each capacitor.

It is given that,

$$\begin{gathered} \mathrm{q}=\mathrm{CV} \\ \mathrm{C}=\frac{\mathrm{q}}{\mathrm{V}} \end{gathered}$$

As the vertical scale set is by ${\mathrm{q}}_{\mathrm{s}}=16\mathrm{\mu C}$and the horizontal scale is set by ${\mathrm{V}}_{\mathrm{s}}=2\mathrm{V}$.

Consider two points on the line 1:role="math" localid="1661753003473" ${\mathrm{P}}_1\left(1,6\right)\mathrm{and}{\mathrm{P}}_2\left(2,12\right)$

$$\begin{gathered} {\mathrm{C}}_1=\frac{{\mathrm{y}}_2-{\mathrm{y}}_1}{{\mathrm{x}}_2-{\mathrm{x}}_1} \\ {\mathrm{C}}_1=\frac{12\mathrm{\mu C}-6\mathrm{\mu C}}{2\mathrm{V}-1\mathrm{V}} \\ {\mathrm{C}}_1=6\mathrm{\mu F} \end{gathered}$$

Now, consider two points on the line 2:${\mathrm{P}}_1\left(1,4\right)\mathrm{and}{\mathrm{P}}_2\left(2,8\right)$

$$\begin{gathered} {\mathrm{C}}_2=\frac{8\mathrm{\mu C}-4\mathrm{\mu C}}{2\mathrm{V}-1\mathrm{V}} \\ {\mathrm{C}}_2=4\mathrm{\mu F} \end{gathered}$$

Similarly, consider two points on the line 3: ${\mathrm{P}}_1\left(1,2\right)\mathrm{and}{\mathrm{P}}_2\left(2,4\right)$

$$\begin{gathered} {\mathrm{C}}_3=\frac{4\mathrm{\mu C}-2\mathrm{\mu C}}{2\mathrm{V}-1\mathrm{V}} \\ {\mathrm{C}}_3=2\mathrm{\mu F} \end{gathered}$$

Using the formulae for the equivalent capacitance of series and parallel combinations, the total equivalent capacitance is given by,

$\frac{1}{{\mathrm{C}}_{\mathrm{eq}}}=\frac{1}{{\mathrm{C}}_1}+\frac{1}{{\mathrm{C}}_2+{\mathrm{C}}_3}$

Substituting the values,

$$\begin{gathered} \frac{1}{{\mathrm{C}}_{\mathrm{eq}}}=\frac{1}{6\mathrm{\mu F}}+\frac{1}{4\mathrm{\mu F}+2\mathrm{\mu F}} \\ \frac{1}{{\mathrm{C}}_{\mathrm{eq}}}=\frac{1}{6\mathrm{\mu F}}+\frac{1}{6\mathrm{\mu F}} \\ \frac{1}{{\mathrm{C}}_{\mathrm{eq}}}=\frac{12}{36\mathrm{\mu F}} \\ {\mathrm{C}}_{\mathrm{eq}}=3\mathrm{\mu F} \end{gathered}$$

The voltage across the capacitor is,

$$\begin{gathered} {\mathrm{V}}_1=\frac{{\mathrm{C}}_{\mathrm{eq}}\mathrm{V}}{{\mathrm{C}}_2+{\mathrm{C}}_3} \\ {\mathrm{V}}_1=\frac{3\mathrm{\mu F}\times 6\mathrm{V}}{4\mathrm{\mu F}+2\mathrm{\mu F}} \end{gathered}$$

The voltage across the capacitor 2 is,

$$\begin{gathered} {\mathrm{V}}_2=\mathrm{V}-{\mathrm{V}}_1 \\ =6\mathrm{V}-3\mathrm{V} \\ =3\mathrm{V} \end{gathered}$$

So, the charge on the capacitor 2 is given by,

$$\begin{gathered} {\mathrm{q}}_2={\mathrm{C}}_2{\mathrm{V}}_2 \\ =4\mathrm{\mu F}\times 3\mathrm{V} \\ =12\mathrm{\mu C} \end{gathered}$$

Hence, the charge stored on the capacitor 2 is${\mathrm{q}}_2=12\mathrm{\mu C}$

Therefore, by using the concept of capacitance, we found the charge stored on the capacitor 2.