Question

If two similar large plates each of area having surface charge density is +a and -b are separated by a distance d in air find the expression for the potential difference and capacitance between them

Answer 1

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