Question

find the electric potential at a point due to a positive charge of 100 microcoulomb at a distance of 5 metre

Answer 1

V = 9*10^9*10^-4/5
V = 1.8*10^5 Volts

Answer 2

Heya friend!!!

As we know that :-

$= > v = \frac{1}{4\pi.e _{0}k } . \frac{q}{r} \\ \\ = > v = 9 \times {10}^{9} \times \frac{100 \times {10}^{ - 6} }{5 } \\ \\ = > v = \frac{9}{5} \times {10}^{5} v\\ \\ = > v = 1.8 \times {10}^{5} v$
___________________[ANSWER]

====================================

_-_-_-_☆$BE \: \: \: \: BRAINLY$☆_-_-_-_