Electrostatic, 3 point charges

jfierro

This is from Sadiku's Elements of electromagnetics. I have come to a result but it's different from that of the book.

1. The problem statement, all variables and given/known data

Three identical small spheres of mass m are suspended by threads of negligible
masses and equal length l from a common point. A charge Q is divided equally
between the spheres and they come to equilibrium at the corners of a horizontal equilateral
triangle whose sides are d. Show that

[tex]Q^2 = 12\pi\epsilon_0mgd^3 \begin{bmatrix} l^2 - \frac{d^2}{3}\end{bmatrix}^{-\frac{1}{2}}[/tex]


2. Relevant equations

[tex]F = \frac{q_1q_2}{4\pi\epsilon_0R^2}\mathbf{a_r}[/tex]

This is all given by the book.

3. The attempt at a solution

The way I see this problem is as follows:



P being the common point, Fe being the force exerted by the other 2 speheres, mg being the force exerted by gravity.
The equilateral triangle would be the top view, while the right triangle is a side view of one of the spheres.

The length of w in the right triangle is:

[tex]d\sqrt{3}[/tex]

from the inscribed circle formula in an equilateral triangle.

If we define alpha as the P angle on the right-angle triangle, and T as the tension on each thread caused by both the electrostatic force and weight of a sphere, then:

[tex]T\sin \alpha = F_e[/tex]

[tex]T\cos \alpha = mg [/tex]

[tex]\frac{\sin \alpha}{\cos \alpha} = \frac{F_e}{mg} (1)[/tex]

But, by the superposition principle:

[tex]F_e = \frac{1}{4\pi\epsilon_0} \frac{q^2}{d^2} + \frac{1}{4\pi\epsilon_0} \frac{q^2}{d^2} = \frac{1}{2\pi\epsilon_0} \frac{q^2}{d^2} (2) [/tex]

where

[tex]q = \frac{Q}{3} (3)[/tex]

is the charge of an individual sphere.

Now,

[tex]\tan \alpha = \frac{w}{h}[/tex]

[tex]h = \frac{w}{\tan \alpha}[/tex]

[tex]l^2 = \frac{w^2}{\tan^2 \alpha} + w^2 = \frac{3d^2}{\tan^2 \alpha} + 3d^2[/tex]

[tex]\tan^2 \alpha = \frac{3d^2}{l^2 - 3d^2}[/tex]

[tex]\tan \alpha = \sqrt{3}d( l^2 - 3d^2 )^{-\frac{1}{2}} (4)[/tex]

Substituting (2), (3) and (4) in (1) and solving for Q yields:

[tex]Q^2 = 18\sqrt{3}\pi\epsilon_0mgd^3( l^2 - 3d^2 )^{-\frac{1}{2}}[/tex]

What's wrong?

Thanks and best regards.

willem2

No time to read it all, but you should have [itex] d = \sqrt{3} w [/itex] and not [itex] w = \sqrt{3} d [/itex]

inti

not understand how you get d=(3)^(1/2)

inti

w=d(3)^(1/2)

saadqureshi

w is not d*(3^1/2)
its d/(3^1/2)

in the picture youre looking at the pyramid from the top..

Attached Thumbnails

 

inti

got the solution....thanks

saadqureshi

Again looking at the pyramid from the top we see that the horizontal component of the force due to the two charges gets cancelled out...

Attached Thumbnails

   

saadqureshi

in the end use the followin equation

tsin a/tcos a= Fe/mg

inti

gr8 work qureshi......

saadqureshi

thank you inti

usamaabdullah

Thanks, I got my answer..

Foxhound31

Suggestion, its better to use (I) and (IV) in equation labelling than (1) (4) for it looks like it is a scalar multiplication