91Ó°ÊÓ

The energy storage in the capacitor is related to the potential difference by the equation in the form

$∆U=\frac{1}{2}C{V}^2$
$∆U=\frac{1}{2}\frac{{\mathrm{Î\mu }}_{0}A}{d}{V}^2$ (1)

Where $C=\frac{{\mathrm{Î\mu }}_{0}A}{d}$. In the first experiment when the battery is kept connected, the voltage will be constant So the data in the d graph is set to the first experiment where V, A and ${\mathrm{Î\mu }}_0$are constant which keeps the graph to be linear. While for the second experiment the voltage changes as the distance changes, so the graph will be non-linear

We could calculate the area of the plate by using equation (1) by taking two points on the graph and getting the slope where the slope will be

$Slope=0.5{\mathrm{Î\mu }}_{0}A{V}^2$

Where V is constant Now we can plug our values into this equation to get the area

$$\begin{gathered} \frac{\left(50-30\right)×{10}^{-9}J}{\left(13-5\right)×{10}^2{m}^{-1}}=0.5{\mathrm{Î\mu }}_{0}A{V}^2 \\ 3.75×{10}^{-11}J.m=0.5{\mathrm{Î\mu }}_{0}A{V}^2 \end{gathered}$$ (2)

Now we can solve equation (2) for A and put our values for V and, to get A

$$\begin{gathered} A=\frac{3.75×{10}^{-11}J.m}{0.5{\mathrm{Î\mu }}_{0}A{V}^2} \\ A=\frac{3.75×{10}^{-11}J.m}{0.5\left(8.854×{10}^{-12}{C}^2/N{m}^2\right){\left(24.0V\right)}^2} \\ A=147c{m}^2 \end{gathered}$$

As shown by equation (1), as the distance d increases, the energy decreases. So in the first experiment, the voltage is constant, which means the energy will decrease but in the second experiment, the voltage will increase as d decreases to keep the energy high. So the battery is disconnected, and there is more energy stored in the capacitor