Question

Fig. 25-33 shows a circuit section of four air-filled capacitors that is connected to a larger circuit. The graph below the section shows the electric potential V(x)as a function of position xalong the lower part of the section, through capacitor 4. Similarly, the graph above the section shows the electric potential V(x)as a function of position xalong the upper part of the section, through capacitors 1, 2, and 3. Capacitor 3 has a capacitance of$\mathbf{0}\mathbf{.}\mathbf{80}\mathbf{}\mathbf{\mu F}$What are the capacitances of (a) capacitor 1 and (b) capacitor 2?

Short answer

a) The capacitance of the capacitor 1 is${\mathrm{C}}_1=2\mathrm{\mu F}$

b) The capacitance of the capacitor 2 is${\mathrm{C}}_2=0.80\mathrm{\mu F}$

Step-by-step solution

Step 1: Given Data

The graph of electric potential V(x) versus position x.

From the graph, the total potential is 12 V,$V_1=2V$and ${\mathrm{V}}_2=5\mathrm{V}$

$C_3=0.80\mathrm{\mu F}$

Determining the concept

From the graph, find the voltage across the capacitor 3. Also, find the total charge on the capacitor 3by using the concept of capacitance.When the capacitors are in a series, they contain the same charge.So easily find the capacitance of capacitors1 and 2.

Formulae are as follows:

q =CV

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

(a) Determining the capacitance of the capacitor

Find the voltage across the capacitor 3 from the graph.

The total potential from the graph is 12 V, so the potential drop across the capacitor 3 is,

$$\begin{gathered} {\mathrm{V}}_3=12\mathrm{V}-{\mathrm{V}}_1-{\mathrm{V}}_2 \\ =12-2\mathrm{V}--5\mathrm{V} \\ =5\mathrm{V} \end{gathered}$$

Since q =CV,

So we can find the charge on the capacitor 3.

$$\begin{gathered} {\mathrm{q}}_3={\mathrm{C}}_3{\mathrm{V}}_3 \\ =0.80\mathrm{\mu F}\times 5\mathrm{V} \\ =4\mathrm{\mu C} \end{gathered}$$

When the capacitors are in a series; they contain the same charge.

Therefore,${\mathrm{q}}_1={\mathrm{q}}_2={\mathrm{q}}_3=4\mathrm{\mu C}$

For capacitor 1,

From the graph,$V_1=2V$

Since,$q_1=C_1{V}_1$or,

$$\begin{gathered} {\mathrm{C}}_1=\frac{{\mathrm{q}}_1}{{\mathrm{V}}_1} \\ {\mathrm{C}}_1=\frac{4\mathrm{\mu C}}{2\mathrm{V}} \\ =2\mathrm{\mu F} \end{gathered}$$

Hence, the capacitance of the capacitor 1 is${\mathrm{C}}_1=2\mathrm{\mu F}$

(b) Determining the capacitance of the capacitor 2

For capacitor 2,

From the graph,$V_2=5V$

Since$q_2=C_2{V}_2$or,

$$\begin{gathered} {\mathrm{C}}_2=\frac{{\mathrm{q}}_2}{{\mathrm{V}}_2} \\ {\mathrm{C}}_2=\frac{4\mathrm{\mu C}}{5\mathrm{V}} \\ =0.80\mathrm{\mu F} \end{gathered}$$

Hence, the capacitance of the capacitor 2 is${\mathrm{C}}_2=0.80\mathrm{\mu F}$

Therefore, by using the relation between charge and capacitance, find the capacitance of both the capacitors.