Question

Consider a capacitor made of two rectangular metal plates of length L and width W, with a very small gap s between the plates. There is a charge +Qon one plate and a charge −Qon the other. Assume that the electric field is nearly uniform throughout the gap region and negligibly small outside. Calculate the attractive force that one plate exerts on the other. Remember that one of the plates doesn’t exert a net force on itself

Short answer

The force on the one plate due to another plate is $\frac{Q_1{Q}_2}{2A{\epsilon }_0}$.

Step-by-step solution

Identification of given data

The given data can be listed below,

• The charge on first plate is, $Q_1$
• The charge on second plate is, $Q_2$
• The electric field is uniform throughout the gap.

Concept/Significance of capacitor.

The typical electrical gadgets employ capacitors as part of their electrical circuits. The capacitor's charge capacity is increased by the nonconducting dielectric.

Determination of the attractive force that one plate exerts on the other

The electric field inside the gap between the two disks is due to charges on each plate.

The electric field due to one disk knowing that the gap between the two disk is very small is given by,

$E=\frac{Q}{2A{\epsilon }_0}$

Here, Q is charge on plates, A is the area of plate and${\epsilon }_0$is the permittivity of free space.

The force on the second plate due to first is given by,

$F_1=Q_2{E}_1$

Here,$Q_2$is the charge on second plate and$E_1$is the electric field due to first plate.

Substitute values in the above equation.

$F_1=Q_2\frac{Q_1}{2A{\epsilon }_0}$

The force on first plate due to second plate is given by,

$F_2=-F_1$

The force is same in magnitude but opposite in the direction.

Thus, the force on the one plate due to another plate is $\frac{Q_1{Q}_2}{2A{\epsilon }_0}$.