Calculate the value of electric field intensity due to a charge of 4μC if the force due to the charge is 8N
-
2.0 x 10\(^6\)NC\(^{-1}\)
- 3.0 x 10\(^6\)NC\(^{-1}\)
- 4.0 x 10\(^6\)NC\(^{-1}\)
- 8.0 x 10\(^6\)NC\(^{-1}\)
The correct answer is: A
Explanation
Electric field intensity(E) = \(\frac{\text{Force}}{\text{charge}}\) = \(\frac{F}{q}\)
q = 4μC = 4.0 x 10\(^{-6}\)C
F = 8N
E = \(\frac{8}{4.0 \times 10^{-6}}\) = 2.0 x 10\(^6\)NC\(^{-1}\)
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