Question

A charge of 2.1 nC and a charge of 4.7 nC are separated by 57.90 cm. Find the equilibrium position for a -4.7 nC charge.
____ cm from the 2.1 nC charge.

what is the answer? and how did you get it? thaanks

Answer 1

The equilibrium position is between the two charges,there tha attractive forces to the +2.1nC and +4.7 nC charges are equal and opposite. If x is the distance from the 2.1 nC charge, require that
2.1/x^2 = 4.7/(57.9-x)^2

Solve that equation for x, which will be in centimeters. The -4.7 nC charge cancels out, as does the Coulomb constant, making the solution easy. You can use nC for charge and cm for distance on both sides of the equation.

(57.9-x)/x = sqrt(4.7/2.1) = 1.496

57.9/x = 2.496
x = ?