Name: Energy Density in a Capacitor
Uploaded: 2019-07-19
Duration: 3 min 1 s
Description: The energy in the capacitor is stored in the form of the electric field. Let us derive a mathematical formula to calculate this energy in terms of the electr...

Question

We have a parallel plate air capacitor of separation $d$ and the length and width of the plates are $L$ and $b$ respectively. It is connected to a cell of emf $E$ Fig. A dielectric slab of

mass $m$ has the same dimension as the inner dimensions of the capacitor. We insert the slab into the space between the capacitor slowly upon a distance $l$ and then release it from rest. Show that in absence of friction it oscillates about that position and find the time-period.

Accepted Answer

There will be 2 capacitance in parallel combination. The first capacitance will be due to the part of dielectric slab inserted between the plates and the second will be due to the remaining space not filled by the dielectric slab.

Let the capacitance due to dielectric $= C_{1} = \frac{ε_{\circ} A K}{d} = \frac{ε_{\circ} (l \times b) K}{d}$

Let the capacitance due to the remaining space be $= C_{2} = \frac{ε_{\circ} A K}{d} = \frac{ε_{\circ} (L - l) b}{d}$

Since $C_{1} a n d C_{2}$ are in parallel, Therefore net capacitance $= C_{1} + C_{2} = \frac{ε_{\circ} (l \times b) K}{d} + \frac{ε_{\circ} (L - l) b}{d} = \frac{ε_{\circ} b (L - l + K l)}{d}$

Energy stored in the capacitor is $= U = \frac{1}{2} c V^{2} = \frac{1}{2} C_{n e t} E^{2} = \frac{ε_{\circ} b (L - l + K l)}{2 d} \times E^{2}$

Then the Force exerted inward is given as $= F = |\frac{d U}{d x}| = \frac{ε_{\circ} b E^{2} (K - 1)}{2 d}$

Therefore $a = \frac{ε_{\circ} b E^{2} (K - 1)}{2 d m}$

To acquire the whole area between the capacitor the time taken can be calculated as : $\frac{1}{2} a t^{2} = S \Rightarrow t = \sqrt{\frac{2 (L - l)}{a}}$

Now the velocity will be maximum, so the slab will still move till its velocity is 0.The time taken will be same as the time taken to acquire max velocity as the acceleration is same.

After the velocity is 0.The force applied will be inward and so the slab will move again to acquire maximum velocity and the move further right to make its velocity 0. Now it is at its initial position.This process will keep on happening.

Therefore, the Time period $= T = 4 t = 4 \times \sqrt{\frac{2 (L - l)}{a}} = 4 \times \sqrt{\frac{2 (L - l)}{a}} = 4 \times \sqrt{\frac{2 (L - l)}{\frac{ε_{\circ} b E^{2} (K - 1)}{2 d m}}}$

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