Question

A 5.61μC and a -2.04μC charge are placed 19.0cm apart. Where can a third charge be placed so that it experiences no net force? [Hint: Assume that the negative charge is 19.0cm to the right of the positive charge.]

Answer 1

no net force implies that the repulsion due to positive charge and attraction due to the negative charge should balance each other. Since the charge on the positive charge particle is larger than the charge on the negative charge particle, the third charge should be on the right of the negative charge given that the negative charge is on the right of the positive charge.

let the distance between the third charge and the negative charge be 'x' cm. So, field due to the first two charges should cancel each other at the place where the third charge is placed for it to experience no net force.

so, K(5.61)/(19+x)² + K(-2.04)/x² = 0, where K is the coulomb's constant.

this imples, 5.61x² = 2.04(19+x)², this implies, 3.57x² - 77.52x - 736.44 = 0

solving this, we get, x = 28.862 cm or x = -7.1474 cm

negative value of x is not possible because attraction and repulsion would be in the same direction and this cannot create a net zero force. so distance is 28.862 cm to the right of the negative charge.