Question

What are the magnitude and direction of the electric field at a distance of 1.5 m and 50 nC charge?

Answer 1

The expression for the electric field is -  

E = k q / r^2

where -

k = 9x10^9 in MKS
q= 50 x 10^-9 C = 5 x 10^-8 C
r =1.5m

put the values in the above expression -

E = 9.10^9 x 5x10^8C/(1.5m)^2 = 200N/C

Now, if the charge is positive, the field points away from the charge; if the charge is negative, the field points toward the charge.

Given that the charge is 50 nC, means it is a positive charge, then the field points outward, away from the charge.

So, magnitude = 200 N/C

Direction = Away from the charge.