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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

But that's not the right answer. Can someone PLEASE HELP!

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# Homework Help: Three Charges - Electric Field Question

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