Question

Derive an expression for the electric field intensity at the surface of the plate.​

Answer 1

Answer:

Electric field intensity is the strength of an electric field at any point. It is equal to the electric force per unit charge experienced by a test charge placed at that point, or E=

q

F

From Gauss law, considering a spherical surface with charge at its center as Gaussian surface,

E(A)=

ϵ

0

q

⟹E(4πr

2

)=

ϵ

0

q

⟹E=

4πϵ

0

r

2

q

which is the intensity of electric field at a distance r from the charge.