Calculate the electric potential at a distance of 20.0cm from a point charge of 0.015C placed in air of permittivity (Take \(\frac{1}{4\piεo}\) as 9.0 x 10\(^9\) Nm\(^2\)C\(^2\))
A
3.40x106V
B
6.75x108V
C
6.75x1012V
D
3.40x105V
E
8.30 x 102V
Verified Correct Answer: B.
Physical Principles & Numerical Solution
To calculate the electric potential \( V \) at a distance \( r \) from a point charge \( Q \):
Given: \( Q = 0.015 \, \text{C} \), \( r = 0.20 \, \text{m} \)
\( \frac{1}{4\pi \varepsilon_0} = 9.0 \times 10^9 \, \text{N m}^2/\text{C}^2 \)
The electric potential is given by:
\(V = \frac{1}{4\pi \varepsilon_0} \cdot \frac{Q}{r} = 9.0 \times 10^9 \cdot \frac{0.015}{0.20} = 6.75 \times 10^8 \, \text{V}\)
Thus, the electric potential is: \(6.75 \times 10^8 \, \text{V}\)
Source: Myschool Past Questions Archive (Record #25091)