Two point charges of +49 esu and +81 esu are placed at a separation of 100 cm in air. Determine the position of the neutral point in the electric fields of the two charges. What is the electric potential at the natural point?

Explanation

The neutral point must be in between the two charges because both the charges are positive. Let +49 esu and +81 esu of charges be placed respectively at A and B.

Let the neutral point P be at a distance x cm from the point A. So the distance of P from the point B is (100-x) cm.
According to the question,
`(49)/(x^(2)) = (81)/((100-x)^(2)) or, 7/x = 9/(100-x) or, x = 43.75cm`
So the neutral point is on the line AB at a distance of 43.75 cm from the charge +49 esu.
If V be the potential at the neutral point, then
`V = 49/(43.75) + (81)/(56.25) = 1.12 + 1.44 = 2.56` esu of potential