Answer:
The electric potential is 15.23 V
Explanation:
Given;
charge Q = 8.473 nC = 8.473 x 10⁻⁹ C
position of the charge, r = 5 m
Coulomb constant is 8.98755 × 10⁹ N·m² /C²
Electric potential is given as;
[tex]V= \frac{KQ}{r} = \frac{8.98755 X10^9 X 8.473X10^{-9}}{5}\\\\V = 15.23 V[/tex]
Therefore, a charge of 8.473 nC uniformly distributed along the x-axis from −1 m to 1 m, the electric potential (relative to zero at infinity) of the point at 5 m on the same axis is 15.23 V.
