physics
posted by brittney .
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

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/(57.9x)^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.
(57.9x)/x = sqrt(4.7/2.1) = 1.496
57.9/x = 2.496
x = ?