Question

A small object carrying a charge of - 4 x 10^-9 C experiences a force of 15 x 10^-6 N in the
negative x-direction when placed at a point in
an electric field. Calculate the electric field at
this point. What would the magnitude and
direction of the force acting on a proton placed at this point be ?

Answer 1

The electric field strength

$$E=\frac{F}{q}=\frac{15\times 10^{-6}\:\rm{N}}{4\times 10^{-9}\:\rm{C}}=3750\:\rm{N/C}$$

The magnitude of the force acting on a proton placed at this point would be

$$F=eE=1.6\times 10^{-19}\:\rm{C}\times 3750\:\rm{N/C}=6\times 10^{-16}\:\rm{N}$$

Since the charge of the proton is positive, so the force is directed in the positive x-axis.