Two point charges q1 20 nC and q2 40 nC are placed 100 cm a

Two point charges (q1 = +20 nC and q2 =-40 nC) are placed 10.0 cm apart along the x axis. Where carn 5

Solution

We have to place a positive charge in such a manner that net force on it is zero. Now like charges repel and unlike attract. Further, the magnitude of negative charge is greater than that of positive.

Therefore the Q3 should be placed closer to Q1. We assume that distance to be x m

Net force = 8.99 x 10^9 [20 x 20 / x^2 - 800/ (x+0.1)^2]

This net force should be zero. That is,

(x + 0.1) / x = 1.414

Hence x = 0.1 / 0.414 = 0.242 metres = 24.2 cm