WAEC Physics 2013 Paper

Concept:The electric field intensity is defined as the electric force acting on a unit positive charge placed in the field.Explanation:The formula for electric field intensity is $E = \frac{F}{q}$, where $F$ is the electric force and $q$ is the charge experiencing the force.Here, the force is $F = 80\ \text{N}$ and the charge is $q = 2.0 \times 10^{-5}\ \text{C}$.Substitute the given values into the formula:$E = \frac{80}{2.0 \times 10^{-5}}$Simplify the fraction:$E = \frac{80}{2} \times \frac{1}{10^{-5}} = 40 \times 10^{5}$Therefore, $E = 4.0 \times 10^{6}\ \text{N C}^{-1}$.This matches option B.Answer:The magnitude of the electric field intensity is $4.0 \times 10^{6}\ \text{N C}^{-1}$.Correct option: B.