Question

What would be the potential at a point p due to a charge of 6*10^-7 located 9cm away​

Answer 1

Answer:

The potential at the required point would be 6 x $10^{-14}$

Explanation:

$V = \frac{k Q}{r}$     Q is charge

                 r is distance of the point from the charge

                 k is constant

V = 9 x $10^{-9}$ x 6 x $10^{-7}$ / 9 x $10^{-2}$                 (Convert 9 cm to m)

   = 6 x $10^{-14}$ V