# Math Help - 4 Charged Particles, find net force

1. ## 4 Charged Particles, find net force

I'm very frustrated with this problem, I have no idea what I could be doing wrong.

In fig 21-23, the particles have charges q1 = -q2 = 100nC and q3 = -q4 = 200nC. The distance a=5.0 cm. What are the x and y components of the net force on particle 3?

The figure looks like this:
12
34

A square with all sides 5.0 cm.

Here's what I'm doing..

Since the net force on particle 3 is due to the sum of the forces due to 1, 2, and 4, I first find each of those by first noting the unit vector that connects them to particle 3.

The unit vector for 1 to 3 is -j
The unit vector for 2 to 3 is (-1/sqrt(2))i + (-1/sqrt(2))j
The unit vector for 4 to 3 is -i.

Next, using Coulombs law, I found the magnitude, noting whether or not it should oppose the unit vector above.

My results,

Force on 3 due to particle 1 = -0.07192j
Force on 3 due to particle 2 = 0.0508i + 0.0508j
Force on 3 due to particle 4 = .14384i

Net:

.19464i + -.02112j

Which isn't even close, though the signs are correct and the x component should be greater than the y, according to the answer key. I've worked it 3 times and get the same result every time, and I have no clue what I am doing wrong.

Thank you.

2. ## Re: 4 Charged Particles, find net force

q1 = -q2 = 100nC and q3 = -q4 = 200nC

force of repulsion on 3 by 1 ...

$F_1 = -\frac{(9 \times 10^9)(10^{-7})(2 \times 10^{-7})}{.05^2} j$

force of attraction on 3 by 4

$F_4 = \frac{(9 \times 10^9)(2 \times 10^{-7})(2 \times 10^{-7})}{.05^2} i$

force of attraction on 3 by 2 ...

$F_2 = \frac{(9 \times 10^9)(10^{-7})(2 \times 10^{-7})}{2\sqrt{2} \cdot .05^2} i + \frac{(9 \times 10^9)(10^{-7})(2 \times 10^{-7})}{2\sqrt{2} \cdot .05^2} j$

if I entered all this in my calculator correctly ...

$\sum F = 0.169i - 0.047j$

3. ## Re: 4 Charged Particles, find net force

That's right. Why the 2sqrt(2) in the denomenator for F2? Why isn't it just sqrt(2)?

4. ## Re: 4 Charged Particles, find net force

Duh, I was taking the distance for 2 to 3 to be .05 as well. How silly.

Thank you.