Question

Consider two parallel, oppositely-charged plates, separated by 26.0 cm. The surface charge density on the first plate is σ = -7.28×10^-3 C/m². What potential difference does a charge move through when it travels from the first plate to the second?

Answer 1

The capacitance of a parallel plate capacitor is given by: C = E0*A/d; where E0 is the permittivity of free space, A is the area of the plates and d is the separation between the plates.

Also, C = Q/V; where Q is the magnitude of the charge on each plate and V is the potential difference between the plates.

or, Q/V = E0*A/d

or, V = d/E0*(Q/A)......(1)

Q/A is given as 7.28*10^-3 C/m^2

d is given as 26 cm or 0.26 m

E0 is a constant and equal to 8.85*10^-12 F/m

On solving (1), we gey V = 21.3875*10^7 volt

= 213.875 megavolt Ans.