What's Wrong with My Electrostatic Equilibrium Calculation?

[jfierro]

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

Homework Statement

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

$$Q^2 = 12\pi\epsilon_0mgd^3 \begin{bmatrix} l^2 - \frac{d^2}{3}\end{bmatrix}^{-\frac{1}{2}}$$

Homework Equations

$$F = \frac{q_1q_2}{4\pi\epsilon_0R^2}\mathbf{a_r}$$

This is all given by the book.

The Attempt at a Solution

The way I see this problem is as follows:

http://img256.imageshack.us/img256/6063/sadikuex4.png

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:

$$d\sqrt{3}$$

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:

$$T\sin \alpha = F_e$$

$$T\cos \alpha = mg$$

$$\frac{\sin \alpha}{\cos \alpha} = \frac{F_e}{mg} (1)$$

But, by the superposition principle:

$$F_e = \frac{1}{4\pi\epsilon_0} \frac{q^2}{d^2} + \frac{1}{4\pi\epsilon_0} \frac{q^2}{d^2} = \frac{1}{2\pi\epsilon_0} \frac{q^2}{d^2} (2)$$

where

$$q = \frac{Q}{3} (3)$$

is the charge of an individual sphere.

Now,

$$\tan \alpha = \frac{w}{h}$$

$$h = \frac{w}{\tan \alpha}$$

$$l^2 = \frac{w^2}{\tan^2 \alpha} + w^2 = \frac{3d^2}{\tan^2 \alpha} + 3d^2$$

$$\tan^2 \alpha = \frac{3d^2}{l^2 - 3d^2}$$

$$\tan \alpha = \sqrt{3}d( l^2 - 3d^2 )^{-\frac{1}{2}} (4)$$

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

$$Q^2 = 18\sqrt{3}\pi\epsilon_0mgd^3( l^2 - 3d^2 )^{-\frac{1}{2}}$$

What's wrong?

Thanks and best regards.

 

[willem2]

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

 

[inti]

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

Homework Statement

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

$$Q^2 = 12\pi\epsilon_0mgd^3 \begin{bmatrix} l^2 - \frac{d^2}{3}\end{bmatrix}^{-\frac{1}{2}}$$

Homework Equations

$$F = \frac{q_1q_2}{4\pi\epsilon_0R^2}\mathbf{a_r}$$

This is all given by the book.

The Attempt at a Solution

The way I see this problem is as follows:

[PLAIN]http://img256.imageshack.us/img256/6063/sadikuex4.png

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:

$$d\sqrt{3}$$

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:

$$T\sin \alpha = F_e$$

$$T\cos \alpha = mg$$

$$\frac{\sin \alpha}{\cos \alpha} = \frac{F_e}{mg} (1)$$

But, by the superposition principle:

$$F_e = \frac{1}{4\pi\epsilon_0} \frac{q^2}{d^2} + \frac{1}{4\pi\epsilon_0} \frac{q^2}{d^2} = \frac{1}{2\pi\epsilon_0} \frac{q^2}{d^2} (2)$$

where

$$q = \frac{Q}{3} (3)$$

is the charge of an individual sphere.

Now,

$$\tan \alpha = \frac{w}{h}$$

$$h = \frac{w}{\tan \alpha}$$

$$l^2 = \frac{w^2}{\tan^2 \alpha} + w^2 = \frac{3d^2}{\tan^2 \alpha} + 3d^2$$

$$\tan^2 \alpha = \frac{3d^2}{l^2 - 3d^2}$$

$$\tan \alpha = \sqrt{3}d( l^2 - 3d^2 )^{-\frac{1}{2}} (4)$$

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

$$Q^2 = 18\sqrt{3}\pi\epsilon_0mgd^3( l^2 - 3d^2 )^{-\frac{1}{2}}$$

What's wrong?

Thanks and best regards.

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..

 

[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 canceled out...

 

[saadqureshi]

in the end use the followin equation

tsin a/tcos a= Fe/mg

 

[inti]

in the end use the followin equation

tsin a/tcos a= Fe/mg

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