Homework Help: physics

Posted by J on Monday, January 28, 2008 at 10:23pm.

A point charge Q= 4.60 uC is held fixed at the origin. A second point charge q=1.20 uC with mass of 2.80 * 10^-4 is placed on the x-axis 0.250 m away from the origin. (b) The second point charge is released from rest. What is its speed withn its distance from the origin is (i) .500 m (ii) 5.00 m (iii) 50.0 m

I am not sure how to get the speed from the different distances

physics - drwls, Tuesday, January 29, 2008 at 10:33am
A quick way to do this is to use conservation of energy. The potential energy of the moving charge in the E-field of the fixed charge is:
P.E. = k Q q/R
where k = 8.99*10^9 N m^2/C^2
The kinetic energy of the particle is
(1/2) m V^2

The initial energy, which is all potential, is the total energy, which is constant. Therefore
kQq/(0.25) = kQq/r + (1/2) m V^2

V = sqrt [(2/m)kQq (4 - 1/r)]

Solve for V at each or the r values in your question 1/2, 5 and 50 m.

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