Respuesta :
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
