The plates of planar capacitor are 1m apart, its charge is 1C and voltage is 1V?

Question

What is the force of electrostatic attraction
beween the plates?

Answer 1

If we use the equation C=Q / V, we know the capacitance of the capacitor is C = 1.

If we assume the area is much greater than the distance between the plates then we can say that,

C = Eo * A / d where Eo is the permittivity of free space, A is the area of the plates and d the distance between them. We are told that d = 1 and we calculated that C = 1, therefore the area is given by, A = 1 / Eo.

To find the force between the capacitor it helps to think of a very small change in d, let's call it x. The internal energy of a parallel plate capacitor is known as U = Q*Q / 2*C. This stored energy will change as we increase the distance between the two plates. By conservation of energy we equate this change in internal energy to the work done in moving the plates apart, i.e W = F*x where F is the force between the plates and x is the small displacement of the plates.

We can then say that,

F*x = U_2 - U_1

Now if we consider the capacitor separated from the circuit and as such it's charge will remain constant but the capacitance with change from
C_1 = 1 / d
to
C_2 = 1 / (d + x)

In our equation this gives,
F*x = 0.5 * Q*Q * ( d + x - d)
Rearranging for F this gives,
F = 0.5 N.

Answer 2

You need to know the dimensions of the plates.