# Net Forces of 3 uneqaul charges of equilateral triangle

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1. Feb 23, 2015

### Lena Carling

< Mentor Note -- thread moved to HH from the technical physics forums, so no HH Template is shown >

Three point charges are located at the corners of an equilateral triangle as Q1 in the top, charge Q2 in the left corner and charge Q3 in teh right corner. Each leg of the triangle has a length of 2.03 meters. And the charges have values as follows. Q1 = 4.07 microcoulombs, Q2 = - 7.7 microcoulombs, and Q3 = -6.93 microcoulombs. Calculate the net electric force (both magnitude and direction) exerted on Q1, Q2,Q3.

I was tryting to solve these problem, but did not get proper result.

My solution:

1) For Q1 and Q2 have attrative Forces as Q1 and Q3 , for Q2 and Q3 Forces are repulsive.

2) F1net for Q1:

F1,2 = kQ1Q2/d^2, so F1,2x=-(kQ1Q2/d^2)xcos60; F1,2y= -(kQ1Q2/d^2)xsin60

F1,3 = kQ1Q3/d^2, so F1,3x=(kQ1Q3/d^2)xcos60; F1,3y= -(kQ1Q3/d^2)xsin60

F1x= -(kQ1Q2/d^2)xcos60+(kQ1Q3/d^2)xcos60

F1y= -(kQ1Q2/d^2)xsin60 +kQ1Q3/d^2)xsin60)

I got F1net=square root of ((F1x^2 )+(F1y^2))

But F1x are equal 0 or not? They do not have the same magnitude, but I solve for both cases equal 0 and not, but finally did not get corret answer for F1net magnitude and direction.

3) F2 net for Q2:

F2,1=kQ2Q1/d^2, so F2,1x=kQ2,Q1/d^2xcos60; F2,1y= (kQ2,Q1/d^2)xsin60

F2,3=kQ2Q3/d^2, so F2,3x=-kQ2Q2/d^2; F2,3y=0

F2x=kQ2Q1/d^2xcos60+(-kQ2,3Q2/d^2)

F2y=(kQ2,3Q1/d^2)xsin60

F2net=square root of ((F2x^2 )+(F2y^2))

I and did not get correct answer after calculation for magnitude and direction for F2net.

4) F3 net for Q3:

F3,1=kQ3Q1/d^2, so F3,1x= -kQ3Q1/d^2xcos60; F3,1y=(kQ2,3Q1/d^2)xsin 60

F3,2= kQ3Q2/d^2, so F3,2x=kQ3,2/d^2; F3,2y=0

F3x=-kQ3Q1/d^2xcos60+F3,2x=kQ3,2/d^2

F3y=kQ3,2/d^2

F3net=square root of ((F3x)^2+(F3y)^2)

I did not get right answer after calculation of net F3 and direction too.

I feel that I put correct vectors for x and y for the Forces for all 3 charges, only not sure just about F1x. Please, help me to get right answer, maybe I got mistake in calculations? Thanks. Due is Feb. 25, 2015.

Last edited by a moderator: Feb 23, 2015
2. Feb 23, 2015

### nasu

It looks OK. At least for the first force. I did not read past that.
No, why would you think that Fx is zero?
I cannot tell more unless you show your actual calculation. I think you should get the right answer from these formulas.

3. Feb 23, 2015

### Lena Carling

Thanks, I decided that magnitudes are not equal, so F1x is not equal 0. But when I was plugging numbers finally, I was getting wrong answers few times and got confused about everything in this problem.

4. Feb 23, 2015

### nasu

So, now you've got the right values?

5. Feb 25, 2015

Yes, I did.