Calculate the value of electric field intensity due to a charge of 4μC if the force due to the charge is 8N
A
2.0 x 10\(^6\)NC\(^{-1}\)
B
3.0 x 10\(^6\)NC\(^{-1}\)
C
4.0 x 10\(^6\)NC\(^{-1}\)
D
8.0 x 10\(^6\)NC\(^{-1}\)
Verified Correct Answer: A.
Physical Principles & Numerical Solution
Electric field intensity(E) = \(\frac{\text{Force}}{\text{charge}}\) = \(\frac{F}{q}\)
q = 4\muC = 4.0 x 10\(^{-6}\)C
F = 8N
E = \(\frac{8}{4.0 \times 10^{-6}}\) = 2.0 x 10\(^6\)NC\(^{-1}\)
Source: WAEC / JAMB / NECO Classroom Archive (Record #70680)