Net Electric Force on Charged Spheres


by rum2563
Tags: charged, electric, force, spheres
rum2563
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#1
Jul16-07, 06:23 PM
P: 103
1. The problem statement, all variables and given/known data
Three spheres, each with a negative charge of 4.0 X 10^6 C, are fixed at the vertices of an equilateral triangle whose sides are 0.20 m long. Calculate the magnitude and direction of the net force on each sphere.


2. Relevant equations
Fe = kq1q2 / r^2


3. The attempt at a solution

I used vector components to try to solve this question.

I think that we only have to find the net force on one of the spheres since the radius and the charges are same throughout the system. Sphere 1 is the "main" sphere from which I want to find out the net force.

Force of Sphere 2 on 1:
F2 = kq1q2 / r^2
= (9 X 10^9)(4.0 X 10^-6)^2 / (0.2)^2
= 3.6 N

For the x-component of force of sphere 2 on 1:

F2x = 3.6 X sin 30
= 1.8 N

For the y-component of force of sphere 2 on 1:

F2y = 3.6 X cos30
= 3.12 N


The force that was exerted by sphere 3 on 1 is the same force as 3.6 N since all the values are the same. Only the direction is different.

Here is the vector sum:

x = -3.6 - 1.8 = -5.4
y = 3.12

c^2 = 3.12 ^2 + (-5.4)^2
= 6.2365
= 6.2 N

Therefore, the net force is 6.2 N.

BUT..... how do I get the angle Θ, because in the book it says 150 away from each side.


Please anyone help. It's just a little problem. Thanks.
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Doc Al
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#2
Jul16-07, 06:28 PM
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Quote Quote by rum2563 View Post
BUT..... how do I get the angle Θ, because in the book it says 150 away from each side.
The symmetry of the triangular arrangement should allow you to find the direction of the net force immediately--no calculation required.
rum2563
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#3
Jul16-07, 06:32 PM
P: 103
I have attached the picture of the system. Can you please elaborate a bit more please?

I am thinking that I have to subtract 30 from 180 to get 150. But I just don't know how to prove that in explainable terms.

Please help. Thanks.
Attached Thumbnails
test.gif  

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#4
Jul16-07, 06:36 PM
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Net Electric Force on Charged Spheres


Look at your diagram. Along what line must the net force on charge 1 act? (Draw it.) What angle does that net force vector make with a side?
rum2563
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#5
Jul16-07, 06:43 PM
P: 103
lol. I am seriously confused. I am not sure how to draw but by using tan, I get 30 degress.

tan-1 = 3.12 / 5.4
= 30 degrees.

And I think that since the sphere 3 to sphere 1 line is horizontal, the angle which the net force makes with the horizontal line is 30 degrees. Therefore, 180 - 30 which is 150 degrees is the answer.

Am I on the right track? I know you said no calculations involved but this is the only way I am trying to figure this out.
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#6
Jul16-07, 06:44 PM
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You can also just find the angle from the force components that you've already calculated, assuming you've done that correctly. Use a little trig.
rum2563
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#7
Jul16-07, 06:45 PM
P: 103
Ok, well then I think I almost got it right.
Thanks for your help. This is the first time I am studying electric force and I think that's pretty complicated.

Thanks anyways. All the best.
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#8
Jul16-07, 06:59 PM
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Quote Quote by rum2563 View Post
lol. I am seriously confused. I am not sure how to draw but by using tan, I get 30 degress.
That's perfectly fine (I was just suggesting you do that while you were posting this.)

tan-1 = 3.12 / 5.4
= 30 degrees.

And I think that since the sphere 3 to sphere 1 line is horizontal, the angle which the net force makes with the horizontal line is 30 degrees. Therefore, 180 - 30 which is 150 degrees is the answer.

Am I on the right track? I know you said no calculations involved but this is the only way I am trying to figure this out.
All good. But realize that another solution to your trig equation for tan(theta) is theta = 210 degrees: that's equivalent to 30 degrees below the -x-axis or 150 with respect to the side of the triangle.

But try this. By symmetry, you know that the net force must act right smack in the middle between the two other charges. Since you know the angle between the sides is 60 degrees, the net force must make an angle of 30 degrees below the horizontal axis. Which gives you your answer of 150 degrees with respect to the sides with (almost) no calculation.
rum2563
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#9
Jul16-07, 07:36 PM
P: 103
Wow, thanks for your response. It actually made perfect sense to me. I am happy that you told me how it doesn't involve any calculation since 60 degrees must be divided by 2 since the net force vector goes exactly through the middle. Thanks very much.


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