Question

An electric field of ${10}^5\text{N/C}$ points due west at a certain spot. What are the magnitude and direction of the force that acts on a charge of $-5\mu C$ at this spot ?

• A. $0.2\text{N}$ due east
• B. $0.5\text{N}$ due east
• C. $0.2\text{N}$ due west
• D. $0.5\text{N}$ due west

Answer 1

The correct option is B $0.5\text{N}$ due east

Given that
$E={10}^5\text{N/C}$ due west
$q=-5\mu C$
Force on any charge $q$ is given by,
$F=qE$
$\Rightarrow F=-5\times {10}^{-6}\times {10}^5$
$\Rightarrow F=-0.5\text{N}$ due west
$\Rightarrow F=0.5\text{N}$ due east

$\begin{array}{c}\text{Caution - Always remember that electrostatic force in a positive}\\ \text{charge in an electric field is always the electric field but on a}\\ \text{negative charge, the electrostatic force is opposite to direction}\\ \text{of electric field.}\end{array}$