Answer:
a. V = (a - b)d/2ε₀ b. 2ε₀A/d
Explanation:
a. The potential difference between the plates
Using Gauss' law, we first find the electric field between the plates
ε₀∫E.dA = Q where Q = charge enclosed, E = electric field
Now Q = [a +(-b)]A = (a - b)A where + a and -b are the surface charge densities of the plates and A is the area of the plates.
ε₀∫E.dA = Q
ε₀∫EdAcos0 + ε₀∫EdAcos0 = (a - b)A
ε₀E∫dA + ε₀E∫dA = (a -b)A
ε₀EA + ε₀EA = (a -b)A
2ε₀EA = (a -b)A
E = (a - b)/2ε₀
We now find the potential difference, V between the plates from
V = ∫E.dl
V =E∫dl
V = Ed where ∫dl = d the distance between the plates.
V = (a - b)d/2ε₀
b. The capacitance between them
Capacitance C = Q/V
= (a - b)A ÷ (a - b)d/2ε₀
= 2ε₀A/d
