# Homework Help: Three Charges - Electric Field Question

1. Jan 31, 2010

### r34racer01

Two charges Qc and -Qc (Qc = 7 µC) are fixed on the x-axis at x = -6 cm and x = 6 cm, respectively. A third charge Qb = 1 µC is fixed at the origin.

A particle with charge q = 0.3 µC and mass m = 6 g is placed on the y-axis at y = 10 cm and released. There is no gravity.

(a) Calculate the initial acceleration of the particle.
ax = 238.19m/s^2
ay = 44.95m/s^2
(b) What was the magnitude of the net electric force on q at its point of release?
F = ?N

F = kQq/r^2 E = F/q = KQ/r^2

Ok so the first part wasn't bad. I just figured out what directions the Coulomb force vectors would be in for eack force, and then resolved thos into their x and y components, summed them and then set them equal to m*a and got
ax = 238.19m/s^2 and ay = 44.95m/s^2

Now part b is where I'm getting really stuck. The electric field vectors point in the same direction as the coulomb force vectors, so I basically did the same as above.

E(Qb) = k(1e-6) / 0.1^2 = 899000
E(Qc) = k(7e-6) / 0.1166^2 = 4628717.02
E(Qc)x = 2381887.8 E(Qc)y = 3968832.6
E(Q-c) = k(-7e-6) / 0.1166^2 = -4628717.02
E(Q-c)x = 2381887.8 E(Q-c)y = -3968832.6

The The Qc y components should cancel out and its x components will basically double so then the answer should be F = sqrt( 899000^2 + (2*2381887.8)^2) = 4847859.7N

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2. Jan 31, 2010

### Staff: Mentor

In your solution to part a, you must have first calculated the force components before computing the acceleration. Why not use those results to find the magnitude of the force? (Note that part b asks for the electric force, not the electric field.)

3. Jan 31, 2010

### r34racer01

Oh man I can't believe I made that mistake, that means I had the answer days ago. Thanks you so much Doc Al, I would have never noticed that, the answer turned out to be 1.4543N if anyone was wondering.