Question

The particle P shown in figure has a mass of 10 mg and a charge of −0⋅01 µC. Each plate has a surface area 100 cm² on one side. What potential difference V should be applied to the combination to hold the particle P in equilibrium?

Answer 1

The particle is balanced when the electrical force on it is balanced by its weight.

Thus,

`mg = qE`

`mg = q xx V^'/d.............(1)`

Here,

d = Separation between the plates of the capacitor

V' = Potential difference across the capacitor containing the particle

We know that the capacitance of a capacitor is given by

`C = (∈_0A)/d`

`⇒ d = (∈_0A)/c`

Thus, eq. (1) becomes

`mg = q xx V^' xx C/(∈_0A)`

`⇒ V^' = (mg∈_0A)/(q xx c)`

`⇒ V^' = (10^-6 xx 9.8 xx (8.85 xx 10^-12) xx (100 xx 10^-4))/((0.01 xx 10^-6) xx (0.04 xx 10^-6)`

`⇒ V^' = 21.68  "mV"`

Since the values of both the capacitors are the same,

`V = 2V = 2 xx 21.86 ≈ 43  "mV"`