# Coulomb's law application

Three identical small spheres of mass 'm' are suspended by threads of negligible masses and length'l' from a common point.A charge'q' is equally b/w the spheres and they come to equilibrium at the corners of a horizontal equilateral triangle of side 'a'.Show that:

q^2=12πεmg a^3 [ l^2-(a^2/3) ]^(-.5)

 ebaines Posts: 10,576, Reputation: 5794 Expert #2 Aug 20, 2009, 08:15 AM
I can help you set the problem up, but you're going to have to do the work to get the full solution yourself.

1. Pick one of the spheres and express the electrostatic force acting on it, in the direction that causes the sphere to swing out against gravity. Call this sphere 'A," and the other two spheres we can call 'B' and 'C.' What you have is that the component of electrostatic force from sphere B that causes sphere A to swings outward is sqrt(3)/2 times the direct elecrostatic force between A and B. If you apply Coloumb's law you can come up with and expression for this. Same with the repelling force from sphere C acting on A. Add the two forces together, and you have the total force that is pushing sphere A outward against gravity.

2. Deteremine the force of gravity that is causing A to want to swing back inward. If you draw a diagram you'll see that the force of gravity is F=mg downward, and the component of mg that causes sphere A to want to move back to the center is mg*tan(theta), where theta is the angle of the string from the vertical. You then have to express tan(theta) in terms of the lengths 'a' and 'l.' Takes a bit of geometry, but not too difficult.

3. Set these two forces equal, and then chug away.

Post back if you're still having difficulty.
 radiation Posts: 21, Reputation: 10 New Member #3 Aug 20, 2009, 10:06 PM
Thank you..I got the solutuion..
 radiation Posts: 21, Reputation: 10 New Member #4 Aug 20, 2009, 10:06 PM
Thank you..I got the solutuion..
 Unknown008 Posts: 8,147, Reputation: 3745 Uber Member #5 Aug 21, 2009, 01:30 AM
Ok, now, I'd like to know how you did this... The force due to the other two charges is $\sqrt3 F$ where F is the force due to one charge only.

Then, net force on the particular charge is 0. The force due to gravity is mg. The total force on the thread is $mg\, tan \theta$

$tan \theta = \frac{\sqrt{\frac{a^2}{3}}}{\sqrt{l^2 - \frac{a^2}{3}}}$

Then I don't know what to do
 ebaines Posts: 10,576, Reputation: 5794 Expert #6 Aug 21, 2009, 06:53 AM
The force acting on a spehere due to gravity that causes it to want to swing back to the center is $mg\cdot tan( \theta )$.

[ EDIT - I Originally had sin (theta) - should be tan (theta) ]

The combined electrostatic force from spheres B and C acting on A in the direction that A swings out is:

$
\sqrt 3 \frac {k_e q^2} {a^2}
$

Where

$
K _e= \frac 1 {4 \pi \epsilon _0 }
$

The $\sqrt 3$ factor comes from the fact that the force from sphere B acting on A has magntude in the outward direction of $\sqrt 3 /2$ times the electrosatic force between A and B , and 1/2 in the perpendicular direction. Same with sphere C. So the total of forces from both spere B and C is $\sqrt 3$ times the electrostatic force.

Set these two forces equal.
 Unknown008 Posts: 8,147, Reputation: 3745 Uber Member #7 Aug 21, 2009, 11:05 AM
Ok, that's what I did:

$F_{gravity} = mg\,tan \theta$

$tan \theta = \frac{\sqrt{\frac{a^2}{3}}}{\sqrt{l^2 - \frac{a^2}{3}}}$

So,

$F_{gravity} = mg $$\frac{\sqrt{\frac{a^2}{3}}}{\sqrt{l^2 - \frac{a^2}{3}}}$$$

$F_{charge} = \sqrt3 F$

$F_{charge} = \sqrt3 $$\frac{q^2}{4 \pi \epsilon_o a^2}$$$

$F_{gravity} = F_{charge}$

$mg $$\frac{\sqrt{\frac{a^2}{3}}}{\sqrt{l^2 - \frac{a^2}{3}}}$$ = \sqrt3 $$\frac{q^2}{4 \pi \epsilon_o a^2}$$$

$\sqrt3 q^2 = \frac{4 \pi \epsilon_o mg a^2 \frac{\sqrt{a^2}}{\sqrt3}}{\sqrt{l^2 - \frac{a^2}{3}}$

$\sqrt3 q^2 = \frac{\frac{4 \pi \epsilon_o mg a^3}{\sqrt3} }{\sqrt{l^2 - \frac{a^2}{3}}$

$q^2 = \frac{4 \pi \epsilon_o mg a^3}{3 \sqrt{l^2 - \frac{a^2}{3}}$

Now, that's not it
 ebaines Posts: 10,576, Reputation: 5794 Expert #8 Aug 21, 2009, 02:26 PM
Quote:
 Originally Posted by Unknown008 Now, that's not it
Hmmm .. I think you're right - the answer as given by the OP seems to be off by a factor of 9.
 radiation Posts: 21, Reputation: 10 New Member #9 Aug 25, 2009, 09:35 AM
The factor of 9 can be accounted for if you consider that Q is distributed among 3 spheres equally..this implies that value of 'q'is Q^2/9 rite..
 Unknown008 Posts: 8,147, Reputation: 3745 Uber Member #10 Aug 25, 2009, 09:54 AM
Well, I don't know.... I have just started with electrostatic forces at school, and seeing this question was like a 'further' question, or some sort of challenge for me... However, I don't understand even now...

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