Question

The potential at a point situated at a distance of 50 cm from a charge of 5 micro Coulomb is what? And its answer is 9E-4 How..????

Answer 1

The answer should be 9E4 (9 × 10⁴) V.

Potential is kQ/r.

k (SI) = $9 * 10^9$

Q = $5 * 10^{-6}$ C

r = $\frac{1}{2}$ m

So, potential = $\frac{9 * 10^9 * 5 * 10^{-6}}{\frac{1}{2}}$ V

= $9 * 10^4$ V

Answer 2

v=kq/r

v=9×10^9×5×10^-6/0.05