Electrostatic, 3 point charges

In summary, the conversation is about solving a problem from Sadiku's Elements of Electromagnetics regarding three identical spheres of mass m suspended by threads from a common point. The spheres are charged equally and come to equilibrium at the corners of an equilateral triangle. The conversation includes equations, diagrams, and a solution attempt. The final solution is given and a suggestion is made for labeling equations.
  • #1
jfierro
20
1
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

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


Homework Equations



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

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 [Broken]

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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  • #2
No time to read it all, but you should have [itex] d = \sqrt{3} w [/itex] and not [itex] w = \sqrt{3} d [/itex]
 
  • #3
jfierro said:
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

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


Homework Equations



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

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 [Broken]

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)
 
Last edited by a moderator:
  • #4
w=d(3)^(1/2)
 
  • #5
w is not d*(3^1/2)
its d/(3^1/2)

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

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  • #6
got the solution...thanks
 
Last edited:
  • #7
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...
 

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  • #8
in the end use the followin equation

tsin a/tcos a= Fe/mg
 
  • #9

saadqureshi said:
in the end use the followin equation

tsin a/tcos a= Fe/mg
gr8 work qureshi...
 
  • #10
thank you inti
 
  • #11
Thanks, I got my answer..
 
  • Like
Likes UDULA
  • #12
Suggestion, its better to use (I) and (IV) in equation labelling than (1) (4) for it looks like it is a scalar multiplication
 

What is electrostatics?

Electrostatics is the study of electric charges at rest. It deals with the behavior of stationary or slow-moving electric charges and how they interact with each other.

What are 3 point charges?

Three point charges refer to three objects with electric charges that are located at distinct points in space. These charges can be positive or negative and can interact with each other through electrostatic forces.

How do 3 point charges interact with each other?

Three point charges interact with each other through electrostatic forces, which can either be attractive or repulsive. The strength of the force depends on the magnitude and distance between the charges.

What is Coulomb's law and how is it related to 3 point charges?

Coulomb's law states that the force between two point charges is directly proportional to the product of their charges and inversely proportional to the square of the distance between them. This law applies to 3 point charges as it helps determine the magnitude and direction of the electrostatic forces between them.

How can 3 point charges be arranged to achieve equilibrium?

In order to achieve equilibrium, the three point charges must be arranged in such a way that the net electrostatic force on each charge is equal to zero. This can be achieved by placing the charges at specific distances and orientations from each other.

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