A charge of 2.9 nC and a charge of 4.6 nC are separated by 40.0 cm. Find the equilibrium position?

Question

A charge of 2.9 nC and a charge of 4.6 nC are separated by 40.0 cm. Find the equilibrium position for a -6 nC charge in meters.
___m (from the 2.9 nC charge)

Answer 1

The quadratic equation you want to solve for x is

6(4.6) x^2 = (1600 - 80x + x^2)(2.9)(6)

The x units will be CENTIMETERS from the 2.9 charge TOWARD the other positive charge. (You can change the centimeters to meters by multiplying by .01.)

You're using Coulomb's law and equating the 2 attractive forces that the positive charges (separated by 40 cm) have on the one negative charge placed somewhere in between. Kq1q3/x^2 = Kq2q3/[(40-x)^2]

q1 = 2.9
q2 = 4.6
q3 = 6 (the neg charge) K will divide out.

(After many careless sign mistakes in using the quadratic formula, I came up with x = 17.703446436469055479839394018196 CENTIMETERS,
which is a ridiculous degree of precision, but you can truncate where YOU want. I didn't round at any time, except when I cked back in the original equation; and the # seems to ck out. (You discard the -154.17 answer. Here I did truncate.))