Electrostatic, 3 point charges


by jfierro
Tags: electrostatic, force, mass
jfierro
jfierro is offline
#1
Nov18-09, 01:17 AM
P: 16
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.
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willem2
willem2 is offline
#2
Nov18-09, 02:08 AM
P: 1,351
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
inti is offline
#3
Mar12-11, 07:34 AM
P: 4
Quote Quote by jfierro View Post
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.
not understand how you get d=(3)^(1/2)

inti
inti is offline
#4
Mar12-11, 07:35 AM
P: 4

Electrostatic, 3 point charges


w=d(3)^(1/2)
saadqureshi
saadqureshi is offline
#5
Mar12-11, 09:43 AM
P: 4
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
Untitled.png  
inti
inti is offline
#6
Mar12-11, 10:38 AM
P: 4
got the solution....thanks
saadqureshi
saadqureshi is offline
#7
Mar12-11, 12:07 PM
P: 4
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
Untitled 2.png   Untitled3.png  
saadqureshi
saadqureshi is offline
#8
Mar12-11, 12:10 PM
P: 4
in the end use the followin equation

tsin a/tcos a= Fe/mg
inti
inti is offline
#9
Mar12-11, 11:15 PM
P: 4
Quote Quote by saadqureshi View Post
in the end use the followin equation

tsin a/tcos a= Fe/mg
gr8 work qureshi......
saadqureshi
saadqureshi is offline
#10
Mar13-11, 03:23 AM
P: 4
thank you inti
usamaabdullah
usamaabdullah is offline
#11
Nov13-11, 12:40 AM
P: 1
Thanks, I got my answer..
Foxhound31
Foxhound31 is offline
#12
Dec30-11, 03:49 AM
P: 1
Suggestion, its better to use (I) and (IV) in equation labelling than (1) (4) for it looks like it is a scalar multiplication


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