Question

Determine the magnitude and direction of the electric field 21.7 cm directly above an isolated 33.0 x 10-5 C charge

Answer 1

The magnitude of the electric field
E=k Q/r^2
E=9×〖10〗^9×(33.0×〖10〗^(-5))/(21.7×〖10〗^(-2) )^2 =6.3×〖10〗^7 V/m=63 MV/m
Since the charge is positive, the field lines point away from it. So for given point electric field would be directed toward up.
Answer: 6.3×〖10〗^7 V/m=63 MV/m