# Electrostatic, 3 point charges

by jfierro
Tags: electrostatic, force, mass
 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 $$Q^2 = 12\pi\epsilon_0mgd^3 \begin{bmatrix} l^2 - \frac{d^2}{3}\end{bmatrix}^{-\frac{1}{2}}$$ 2. Relevant equations $$F = \frac{q_1q_2}{4\pi\epsilon_0R^2}\mathbf{a_r}$$ 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: $$d\sqrt{3}$$ 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.
 P: 1,398 No time to read it all, but you should have $d = \sqrt{3} w$ and not $w = \sqrt{3} d$
P: 4
 Quote by 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 $$Q^2 = 12\pi\epsilon_0mgd^3 \begin{bmatrix} l^2 - \frac{d^2}{3}\end{bmatrix}^{-\frac{1}{2}}$$ 2. Relevant equations $$F = \frac{q_1q_2}{4\pi\epsilon_0R^2}\mathbf{a_r}$$ 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: $$d\sqrt{3}$$ 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)

 P: 4 Electrostatic, 3 point charges w=d(3)^(1/2)
 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
 P: 4 got the solution....thanks
 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
 P: 4 in the end use the followin equation tsin a/tcos a= Fe/mg
P: 4
 Quote by saadqureshi in the end use the followin equation tsin a/tcos a= Fe/mg
gr8 work qureshi......
 P: 4 thank you inti
 P: 1 Thanks, I got my answer..
 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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