EK 1001 Physics # 782

[Zyto]

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Which of the following represents work done on q2 when moved from its present position to a distance r from q1?

E is the electric field created by q1. V is the voltage at a given point in the field, E. Assume that the electric field created by q2 is negligible compared to E. k is Coulomb's constant. m1 is the mass of q1, and m2 is the mass of q2.

The original distance of q2 from q1 is shown in the figure to be 2r.

Therefore, the work done should equal electrical potential energy q1 at 2r minus the electrical potential energy of q1 at r (i.e. the change in potential energy of charge q1).

The electrical potential energy is the strength of the electrical field (kq2/r^2) times the charge of the particle (q1) = (kq1q2/r^2)

= (kq1q2/[2r]^2) - (kq1q2/[r^2)

The book lists the answer as kq1q2/2r but I'm having trouble getting that answer. Could someone tell me where I'm going wrong?

 

[ashtonjam]

In that equation you wrote, you squared the radius on the bottom but that is the force equation, F = kqq/r^2. Potential energy is U = kqq/r.

If Ui = kqq/2r (since q2 is 2r away from q1), and Uf = kqq/r (q2 ends up at a distance of r away from q1)

Change in potential energy = Uf - Ui = kqq/r - kqq/2r = kqq/2r. It's pretty much saying 1 - 1/2 = 1/2.

 

[Zyto]

Ahh, what a dumb mistake.

I see why that is now.

Also work = force * distance, the r/R^2 cancels one of the ones in the denominator.