Question

Two parallel, circular conducting plates 32 cm in diameter are separated by 0.5 cm and have charges of +24 nC and -24 nC, respectively. What is the magnitude of the electric field between these plates?

Answer 1

Answer: E = 33762.39 N/c

Explanation: we calculate the capacitance of the two conducting plates ( this is because, 2 circular disc carrying opposite charges and the same dimension form a capacitor).

C = A/4πkd

Where C = capacitance of capacitor

A = Area of plates = πr² ( where r is radius which is half of the diameter)

K = electric constant = 9×10^9

d = distance between plates = 0.5cm = 0.005 m

Let us get the area, A = πr², where r = D/2 where D = diameter

r = 32/2 = 16cm = 0.16m

A = 22/7 × (0.16)² = 0.0804 m²

By substituting this into the capacitance formula, we have that

C = 0.0804/4×3.142*9×10^9 × 0.005

C = 0.0804/565486677.646

C = 142.17*10^(-12) F.

But C =Q/V where V = Ed

Hence we have that

C = Q/Ed

Where C = capacitance of capacitor = 142.17*10^(-12)F

Q = magnitude of charge on the capacitor = 24×10^-9c

E = strength of electric field =?

d = distance between plates = 0.005m

142.17*10^(-12) = 24 ×10^-9 / E × 0.005

By cross multiplying

142.17*10^(-12) × E × 0.005 = 24 ×10^-9

E = 24 ×10^-9 / 142.17*10^(-12) × 0.005

E = 33762.39 N/c