Name: Force Between Capacitor Plates
Uploaded: 2019-07-19
Duration: 2 min 26 s
Description: In this video, let us derive the expression for the electrostatic force acting between the parallel plates of a capacitor. Watch this video to learn.

Question

Two large parallel metal plates, each having area $A$ , are oriented horizontally and separated by a distance $3 d$ . A grounded conducting wire joins them, and initially each plate carries no charge. Now a third identical plate carrying charge $Q$ is inserted between the two plates, parallel to them and located a distance $d$ from the upper plate, as shown in the figure. What induced charge appears on each of the two original plates?

Accepted Answer

Capacitance of parallel plate capacitor is

$3 d$

Given, There are two large parallel metal plates, each having area $A$ , which are oriented horizontally and separated by a distance $3 d$ . A grounded conducting wire joins them, and initially each plate carries no charge.

Here, the potential of the upper plate and lower plate is zero.

A third identical plate carrying charge $Q$ is inserted between the two plates, parallel to them at a distance $d$ from the upper plate, as shown.

The capacitance of the pair of upper plate and the middle (third) plate is

$C_{1} = \frac{ε_{0} A}{d}$ .

Similarly, the capacitor of the pair of middle plate and the lower plate is

$C_{2} = \frac{ε_{0} A}{2 d}$ .

Thus, $\frac{C_{1}}{C_{2}} = \frac{2 d}{d} = 2$ .

Let's consider the charges that are induced on the surface of each plate as shown in the figure below:

Here, the capacitors $C_{1}$ and $C_{2}$ are connected in parallel. Thus, the potential difference $(∆ V)$ between both the capacitors are same.

Thus, $V = \frac{q}{C_{1}} = \frac{Q - q}{C_{2}}$ .
$\Rightarrow \frac{q}{Q - q} = \frac{C_{1}}{C_{2}} = 2$ or

$\Rightarrow q = \frac{2 Q}{3}$ .
Therefore, the induced charges on the lower surface of upper plate is $- \frac{2 Q}{3}$ and on the upper surface of the lower plate is $- \frac{Q}{3}$ .

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