3 equal charges q are at the vertices of an equilateral triangle as shown, with the z-axis running through the midpoint of the triangle (such that the distance from each charge to the midpoint is d)(adsbygoogle = window.adsbygoogle || []).push({});

a bead of charge Qb (of equal sign as the 3 charges q) is supposed to be levitated on the positive z-azis (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 z-axis

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 z-azis 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 z-component of each radial vector...but how do i do calculate for this z-component?

If i break down the radial force into 2 components, with one in the x-y plane and the other in the z-direction, then the angle etween the xy-plane 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 z-component of the radial vector?

z-component 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...

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# Coulomb's law, balancing a ball with point charges

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