- #1
maCrobo
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Homework Statement
A particle of (positive) charge Q is assumed to have a fixed position at P. A second particle of mass m and (negative) charge -q moves at constant speed in a circle of radius r1, centered at P. Derive an expression for the work W that must be done by an external agent on the second particle in order to increase the radius of the circle of motion, centered at P, to r2.
Homework Equations
W=∆U+∆K where U is the potential energy and K the kinetic energy.
[itex]K=\frac{1}{2}mv^{2}[/itex]
[itex]U=\frac{Q(-q)}{4πε_{0}}\frac{1}{r}[/itex]
[itex]F_{coulomb}=\frac{1}{4πε_{0}} \frac{Q(-q)}{r^{2}}[/itex]
[itex]F_{centrifugal}=m\frac{v^{2}}{r}[/itex]
The Attempt at a Solution
From the fact that [itex]F_{coulomb}=F_{centrifugal}[/itex] I know that v changes as r changes. So I get the two values of v at r1 and r2.
Then I simply wrote W=∆K+∆U explicitly but my result is 3 times the book one. What may be wrong?