Originally Posted by

**Aryth** Well, just think about it, we need the location of a third charge that makes the force double what it currently is on the 1.4nC charge, first we calculate the current force on the 1.4 nC charge:

$\displaystyle F = k\frac{q_1 q_2}{r^2} \approx 1.25 x 10^{-7} N$

So, now we multiply that by 2:

$\displaystyle 2F \approx 2.52 x 10^{-7} N$

Now we need to determine the size of a new charge placed halfway between the two original charges. So the distance is 20cm:

$\displaystyle 2.52 x 10^{-7} N = k\frac{(1.4 x 10^{-9})q_3}{0.2^2}$

$\displaystyle 1.01 x 10^{-8} = k(1.4 x 10^{-9})q_3$

$\displaystyle 0.8 nC = q_3$