# Force on a charge, 4 charges

1. Sep 15, 2008

### rawrlen

1. The problem statement, all variables and given/known data
A charge of 5.00 mC is placed at each corner of a square 0.100 m on a side.

Determine the magnitude of the force on each charge.

2. Relevant equations
F= (kq1q2)/r2

k= 9*10^9

3. The attempt at a solution

I drew 4 charges in a square .1m apart from each other, then used the formula above.

Q1 Q2

Q3 Q4

I did the calculation for Q2 (figuring they all should be the same):

F(2 due to 1) = [(9*10^9)(5*10^-6)^2]/(.01) = 22.5 N
F(2 due to 4) = [(9*10^9)(5*10^-6)^2]/(.01) = 22.5 N
F(2 due to 3) = [(9*10^9)(5*10^-6)^2]/(.02) = 11.25 N

Then:

Ftotal = ($$\sqrt{(22.5^2+22.5^2}$$) + 11.25 = 43.07 N

Did I do this wrong? Or is the distance from each charge not .1 m?

2. Sep 15, 2008

### danago

You seem to have forgotten to square the distances between charges. Also, think again about what the distance between Q2 and Q3 is. The length of the diagonal of a square is not just double the length of each side.

3. Sep 15, 2008

### rawrlen

I did square the dist between charges, .1^2 = .01

You're right Q2 to Q3 is not the double of the length of the sides, I used Pythagorean theorem and squared both sides and took the sqrt of that;

.1^2 + .1^2 = .02

sqrt .02 = .1414

but the dist. needed to be squared for the formula so .1414^2 = .02

lmk if I have mistaken what you posted, and thanks for the reply

4. Sep 15, 2008

### danago

Sorry i read the length of the sides as 0.01m instead of 0.1m; my mistake

5. Sep 15, 2008

### rawrlen

No worries :). Any more suggestions?

6. Sep 16, 2008

### rawrlen

Figured it out..... mC does not mean micro it means MILLA!!!! 10^-3 instead of 10^-6