At what distance from a 1.2 x 10\(^{-7}\)C point charge will the electric field intensity be equal to 4.8 x 10\(^{-4}\)NC\(^{-1}\) [ Take \(\frac{1}{4\pi ε_0}\) = 9.0 x 10\(^9\)]
A
1.5Km
B
3.0Km
C
2.0Km
D
4.5Km
Verified Correct Answer: A.
Physical Principles & Numerical Solution
E = \(\frac{\text{Kq}}{r^2}\)
r = \(\sqrt{\frac{\text{Kq}}{\text{E}}}\)
r = \(\sqrt{\frac{9.0 \times 10^9 \times 1.2 \times 10^{-7}}{4.8 \times 10^{-4}}}\)
r = \(\sqrt{2.25 \times 10^6}\) = 1500m = 1.5km
Source: Myschool Past Questions Archive (Record #74549)