Question

a) What is the magnitude of the electric field 31.8 cm directly above an isolated 2.71×10−5 C charge? b) What is the direction of the electric field? What is the direction of the electric field? upward downward

Answer 1

magnitude of Electric field is given by:

|E| = k*|q|/R^2

Direction of electric field is away from positive charge and towards the negative charge.

Given that:

q = 2.71*10^-5 C

R = 31.8 cm = 0.318 m

So,

|E| = 9*10^9*2.71*10^-5/(0.318^2)

|E| = 2.41*10^6 N/C = 2.41 MN/C

Direction of |E| = Since charge is positive, so electric field will be away from the charge, and since we need electric field at a point above the positive charge, So electric field will be in upward direction