# Help!A question very difficulty

1. Jun 27, 2005

### kidd

Help!A question very difficulty!!

The question is:
Three identical small spheres of mass m are suspended by threads of negligible masses and equal length L from a common point.Acharge Q is divided equally between the spheres and they come to equilibrium at the coners of a horizontal equilateral triangle whose slides are d.

Prove the following Q²=12x∑mgd³[L²-d²/3]^-1/2.

P/S:No figure given,need draw myself!!Any one know it,pls help me.....

2. Jun 27, 2005

### siddharth

If you show what you have done so far, it will be easier to help you and correct mistakes, if any.

3. Jun 27, 2005

### kidd

I don't know how to start to solve this problem!!
i cant understand even how the graph look!T.T
The lecture just throw this question to us.....T.T
PLS HELP ME!!

4. Jun 27, 2005

### Dr.Brain

Ofcourse the charge Q will be distributed equally as Q/3 on all three spheres.Now all three spheres are equally positively charged and will repel each other such that they all repel each other and all three deviate from each other such that , the three strings and the three spheres form edges of a prism, got it? , now they are at equal ditances from each other.Apply the coulombs Law.Show your work.

BJ

5. Jun 27, 2005

### Dr.Brain

To explain in detail , the three spheres will form vertices of the base of prism .

Now visualise that you have three triangles (not the equilateral one but the one formed by strings as two sides and distance 'd' of the equilateral triangle) , Apply Geometry to find out the angles of this triangle.

BJ

6. Jun 27, 2005

### kidd

Is it a triangular prism?

7. Jun 27, 2005

### siddharth

The diagram isn't too difficult. You have three masses hanging from a string and all the strings meet at a common point. It's like a pyramid with a triangular base. At the top of the pyramid is the common point and at each vertex of the triangle at the base, are the spherical masses. Is the diagram clear?

Now Draw the Free-Body Force diagram for ANY ONE spherical mass. What are the forces acting on it?
They are
(i) Gravity
(ii) Tension from the string
(iii) The Electrostatic forces due to each of the other spherical masses.

You also know that the whole system is in equilibirium. So, write the components of the forces in the x,y&z directions and equate them. From this, after some trignometry, you should get the final relation.

8. Jun 27, 2005

### Dr.Brain

Diagram

IN EQUILIBRIUM ALL FORCES (GRAVITY, ELECTROSTATIC AND TENSION) are balanced.

BJ

Something like this:

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9. Jun 27, 2005

### kidd

Thank you all very much!!!!Finally i understand it!!!!!^^
U all so clever!

10. Jul 11, 2005

### kidd

Any one can show me how to get the equation from the graph?and how to prove?Can explain it for me?I use many for it,but still cannot get....

11. Jul 11, 2005

### OlderDan

I'm not sure what your original equation is saying, but I think it should say

$${Q\ }^2 = 12\pi \varepsilon _{\rm{o}} mg\frac{{{d\ }^3 }}{{\sqrt {{L\ }^2 - \frac{{{d\ }^2 }}{3}} }}$$

There are three horizontal forces acting on each charge. Two of them are the repulsive electrical forces that can be found from Coulomb's law. Each charge has magnitude Q/3 and they are separated by distance d. The two repulsions form an angle of 60 degrees, and their sum is offset by the horizontal component of tension in the string. The vertical component of tension must equal the weight of the sphere, mg. The horizontal distance from the veritical line dropped from the point of suspension needs to be found so that you can find the horizontal component of the tension in terms of mg, L and d. There are a couple of ways to do that. One is by recognizing that if you divide the equilateral triangle into 6 congruent right triangles, the hypotnuse of the right triangles is the distance you need. You can find the sides of the right triangle from area considerations.

I'll post a diagram that has all this illustrated, including the dimensions of the triangle formed by the string, the horizontal distance from the vertical line below the point of suspension to the charge and the vertical distance from the point of suspension to the charge. See if you can come up with the value of y in the upper diagram that corresponds to the horizontal distance in the lower diagram. Then do the algebra needed ot get the result.

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