physics
posted by carla .
A charge of 2.0 nC and a charge of 5.3 nC are separated by 41.80 cm. Find the equilibrium position for a 4.7 nC charge. ___ cm from the 2.0 nC charge.

Let x be the distance from the 2.0 nC charge and 41.8 x cm be the distance from the 5.3
For the forces to balance each other,
2.0/x^2 = 5.3/(41.8x)^2
(41.8x)/x = sqrt(5.3/2) = 1.6279
41.8/x  1 = 1.6279
41.8 = 2.6279 x
x = 15.9 cm
41.8 x = 25.9 cm