Question

Basically, I need an explanation for the answer posted, which is a) E=4.495*10^9.

(a) Calculate the strength and direction of the electric field E due to a point charge of +1.00 μC (micro-Coulombs) at a distance of 1.00 mm from the charge.

Answer 1

Expression for the electric field due to a charge ‘q’ at a distance ‘r’ is given as –

E = k*q / r^2 -----------------------------------------(i)

where, k = Coulomb’s constant = 8.99 x 10^9 C^2 / (N*m^-2)

q = 1.0 micro C = 1.0 x 10^-6 C

r = 1.00 mm = 0.001 m = 10^-3 m

Putting the above values in equation (i) –

E = (8.99 x 10^9 x 1.0 x 10^-6) / (10^-3)^2 = 8.99 x 10^9 V/m

I think the answer posted is wrong.

The correct value of the electric field is 8.99 x 10^9 V/m.

Since the charge is positive so the direction of the electric field is away from the charge.