91Ó°ÊÓ

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

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

$C_3=0.80\mathrm{μ¹ó}$

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.

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.

$$\begin{gathered} {\mathrm{q}}_3={\mathrm{C}}_3{\mathrm{V}}_3 \\ =0.80\mathrm{μ¹ó}×5\mathrm{V} \\ =4\mathrm{μ°ä} \end{gathered}$$

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

Therefore,${\mathrm{q}}_1={\mathrm{q}}_2={\mathrm{q}}_3=4\mathrm{μ°ä}$

For capacitor 1,

From the graph,$V_1=2V$

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

$$\begin{gathered} {\mathrm{C}}_1=\frac{{\mathrm{q}}_1}{{\mathrm{V}}_1} \\ {\mathrm{C}}_1=\frac{4\mathrm{μ°ä}}{2\mathrm{V}} \\ =2\mathrm{μ¹ó} \end{gathered}$$

Hence, the capacitance of the capacitor 1 is${\mathrm{C}}_1=2\mathrm{μ¹ó}$

For capacitor 2,

From the graph,$V_2=5V$

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

$$\begin{gathered} {\mathrm{C}}_2=\frac{{\mathrm{q}}_2}{{\mathrm{V}}_2} \\ {\mathrm{C}}_2=\frac{4\mathrm{μ°ä}}{5\mathrm{V}} \\ =0.80\mathrm{μ¹ó} \end{gathered}$$

Hence, the capacitance of the capacitor 2 is${\mathrm{C}}_2=0.80\mathrm{μ¹ó}$

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