Question

A positive charge of 6.0 × 10-4 C is in an electric field that exerts a force of 4.5 × 10-4 N on it. What is the strength of the electric field

Answer 1

The magnitude of the electric force exerted on a charge in an electric field is given by

$F=qE$

where

q is the charge

E is the magnitude of the electric field

In this problem, we have a charge of $q=6.0 \cdot 10^{-4} C$, while the force exerted on it is $F=4.5 \cdot 10^{-4}N$, so we can rearrange the previous formula to calculate the magnitude of the electric field:

$E=\frac{F}{q}=\frac{4.5 \cdot 10^{-4} N}{6.0 \cdot 10^{-4} C}=0.75 N/C$

Answer 2

Field strength = force / charge (E=F/Q)
 E = 4.5×10^-4 / 6.0×10^-4 to obtain .....