91Ó°ÊÓ

1. For capacitor 1, area isA
2. For capacitor 2, area is 2A
3. For capacitor 3, area isA
4. For capacitor 1, separation is d.
5. For capacitor 2, separation isd and capacitor 3, separation is 2d

Using Eq.25-9, we can find the capacitance of each capacitor from the given values of area and plate separation. Then using Eq.25-1, we can compare the capacitance from the slopes of the plots. Then, comparing the predictions about the capacitance yielded from both the equations, we can find out which plot goes with which capacitor.

Formulae:

The charge within the plates of the capacitor,$q=CV$ …(¾±)

The capacitance between the plates with air in the medium,$C=\frac{{∈}_{0}A}{d}$ …(¾±¾±)

From the q-V graph and using equation (i), we can get the capacitance value as follows:

$\begin{array}{l}C=\frac{q}{V}\\ =\mathrm{slope}\end{array}$

Let, the values be A=1 and d=1 for capacitor plates.

The capacitance of capacitor 1 using the given values in equation (ii) can be given as follows:

$$\begin{gathered} C_1={\mathrm{Î\mu }}_0\left(\frac{1}{1}\right) \\ ={\mathrm{Î\mu }}_0 \end{gathered}$$

Similarly, the capacitance of capacitor 2 using the given values in equation (ii) can be given as follows:

$$\begin{gathered} C_2={\mathrm{Î\mu }}_0\left(\frac{2}{1}\right) \\ =2{\mathrm{Î\mu }}_0 \end{gathered}$$

Andthe capacitance of capacitor 3 using the given values in equation (ii) can be given as follows:

$$\begin{gathered} C_2={\mathrm{Î\mu }}_0\left(\frac{1}{2}\right) \\ =\frac{1}{2}{\mathrm{Î\mu }}_0 \end{gathered}$$

From this, we can interpret that capacitance of capacitor 2 has the greatest value and capacitor 1 has the greater capacitance than that of capacitor 3.

From the graph, we can infer that the plot a has the greatest slope. Hence, the capacitance and the slope of plot b is greater than that of plot c. Hence, corresponding capacitance.

Therefore, we can conclude that, Plot a goes with capacitor 2, plot b goes with capacitor 1 and plot c goes with capacitor 3.