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Electrostatic, 3 point charges 
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#1
Nov1809, 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 rightangle 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. 


#2
Nov1809, 02:08 AM

P: 1,396

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
Mar1211, 07:34 AM

P: 4




#4
Mar1211, 07:35 AM

P: 4

Electrostatic, 3 point charges
w=d(3)^(1/2)



#5
Mar1211, 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.. 


#6
Mar1211, 10:38 AM

P: 4

got the solution....thanks



#7
Mar1211, 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...



#8
Mar1211, 12:10 PM

P: 4

in the end use the followin equation
tsin a/tcos a= Fe/mg 


#9
Mar1211, 11:15 PM

P: 4




#10
Mar1311, 03:23 AM

P: 4

thank you inti



#11
Nov1311, 12:40 AM

P: 1

Thanks, I got my answer..



#12
Dec3011, 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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