Coulomb's law, balancing a ball with point charges
3 equal charges q are at the vertices of an equilateral triangle as shown, with the zaxis running through the midpoint of the triangle (such that the distance from each charge to the midpoint is d)
http://img111.imageshack.us/img111/7...phieew7.th.jpg a bead of charge Qb (of equal sign as the 3 charges q) is supposed to be levitated on the positive zazis (coming out of the midpoint of the triangle). Derive an expression for the coloumb force exerted on the bead as a function of its position on the positve zaxis 2. Relevant equations F = (K * q * Qb ) / r^2 is the force in the radial direction (straight line connecting the 2 charges) 3. The attempt at a solution OK I'm kind of having trouble here.... The force from each charge would simply be F = (K * q * Qb ) / (d^2 + z^2) (by pythagorean theorem, the square of the line connecting each charge to any point on the positive zazis would be d^2 + z^2) So far so good because of the way the charges are positioned positioned , the x and y components of the vectors will cancel each other in between the 3 charges (I think) so we're left toworry only about the zcomponent of each radial vector...but how do i do calculate for this zcomponent? If i break down the radial force into 2 components, with one in the xy plane and the other in the zdirection, then the angle etween the xyplane component and the radial component is tan^1 (z/d)....but now hat i have the hypotenuse ((K * q * Qb ) / (d^2 + z^2) ) and the angle tan^1 (z/d), how do i isolate for just the zcomponent of the radial vector? zcomponent of the vector would be the hypotenuse times cos of the angle....but i have the angle expressed as inverse tan of (z/d) so how can i take the cos of something i only have expressed as that? I am totally lost as to how to derive this expression... 
Re: Coulomb's law, balancing a ball with point charges
Radial force can be written as F = i*Fx + j*Fy + k*Fz
Direction of the F is given by direction cosines. z component of the F = Fz = F*cos(gamma)= F*z/r. This is due to one charge. 
Re: Coulomb's law, balancing a ball with point charges
So how does cos(gamma) come out to z/r?

Re: Coulomb's law, balancing a ball with point charges
OH never midn i udnerstand where it comes from....If i do it by similar triangles than
Fz / Ftot = z / r Fz = Ftot * z/r So i guess Fz = kqQb * z / (Z^2+D^2)^2 thanks for the help :) 
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