Calculate the magnitude of the electric field intensity in a vacuum at a distance of 25cm from a charge of 5 × 10-3C. Take (\(\frac{1}{4πε0} = 9\times10^9Nm^2C^{-2}\))
A
7.2 × 108NC-1
B
1.8 × 108NC-1
C
5.6 × 108NC-1
D
3.6 × 108NC-1
E
1.5 × 106NC-1
Verified Correct Answer: A.
Physical Principles & Numerical Solution
Electric field intensity = \(\frac{q}{ 4\piε₀r^2}\) r = 0.25m
= \(\frac{9 \times10^9 \times 5 \times 10^{-3}}{0.25^2}\)
= \(7.20 \times 10^8NC^{-1}\)
Source: WAEC / JAMB / NECO Classroom Archive (Record #42965)