Calculate the electric potential at a distance of 20.0cm from a point charge of 0.015 C placed in air of permittivity. [ take \(\frac{1}{4\pi ε₀}\) = 9.0 x 10\(^9\)Nm\(^2\)C\(^{-2}\)]
A
6.75 x 10\(^6\)V
B
3.40 x 10\(^5\)V
C
3.40 x 10\(^9\)V
D
6.75 x 10\(^8\)V
Verified Correct Answer: D.
Step-by-Step Mathematical Solution
Given: r = 20cm ⇒ 0.20m, K = \(\frac{1}{4\pi ε₀}\) = 9.0 x 10\(^9\)Nm\(^2\)C\(^{-2}\), q = 0.015C
The electrical potential V = \(\frac{\text{kq}}{\text{r}}\)
V = \(\frac{ 9.0 x 10^9 \times 0.015}{0.20}\) = \(\frac{1.35 \times 10^8}{0.20}\) = 6.75 x 10\(^8\)V
Source: WAEC / JAMB / NECO Classroom Archive (Record #76394)