A charge of 3.5 nC and a charge?

Question

Can anyone help me with this question????

A charge of 3.5 nC and a charge of 4.6 nC are separated by 40.0 cm. Find the equilibrium position for a -6 nC charge.

Answer 1

For the purpose of discussion, let us assume that the charge of 3.5 nC is at the origin, and the charge of 4.6 nC is at 40.0 cm along the x-axis.  So the electric field at any point x along the x-axis for x<0 or x>40, is:

E = (1e-9/4πε)*(3.5/x^2 + 4.6/(x-40)^2)...............(1)

and for 0 < x < 40 is:

E = (1e-9/4πε)*(3.5/x^2 - 4.6/(x-40)^2)...............(2)

Let E = 0, we have from (1):

3.5(x-40)^2 + 4.6x^2 = 0

(3.5+4.6)x^2 - 3.5*80x + 3.5*1600 = 0

The roots of the quadratic equation are imaginary.

Let E = 0, we have from (2):

3.5(x-40)^2 - 4.6x^2 = 0

(3.5-4.6)x^2 - 3.5*80x + 3.5*1600 = 0

The roots of the quadratic equation are -273 and 18.6.

Ignore the root -273, since it is outside the range.  So the equilibrium position is 18.6cm from the charge of 3.5 nC for any size of testing charge.