# Math Help - size of a charge

1. ## size of a charge

1. The problem statement, all variables and given/known data
Hi
I need clarification on this problem. Calculate
a) the force between 2 charges of +1.4nC and +1.6nC on point conductors 40cm apart in air.
b)What size of charge on a third point conductor placed midway between the first two conductors would result in doubling of the magnitude of the force on the 1.4nC charge?

2. Relevant equations
Coulombs law= (Kq1q2)/r^2

3. The attempt at a solution
I obtained 1.25 x 10^7N for the first part. For the second part, I stated the new charge as q and the new distance as 20cm. Making the new charge the subject of my formula, i put in the new details. But my result is not correct.

2. 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:

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

So, now we multiply that by 2:

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

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

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

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

$0.8 nC = q_3$

3. 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:

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

So, now we multiply that by 2:

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

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

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

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

$0.8 nC = q_3$
I did exactly what you did. But the book states the answer as 40nC. What do you think?

4. Hmm... No matter how I do it the answer I got keeps popping up... Working backwards (even though it's obvious), you get twice the beginning force... So it should be right, maybe it's a typo.

5. Originally Posted by Aryth
Hmm... No matter how I do it the answer I got keeps popping up... Working backwards (even though it's obvious), you get twice the beginning force... So it should be right, maybe it's a typo.
Possibly. Its not my day. Just not getting anything. Thanks