Electric potential, off by a factor of 1/2

[supersunshine]

Homework Statement

"A particle of positive charge Q is fixed at point p. A second particle of mass m and negative charge -q moves at a constant speed in a circle of radius r1 centered at P. Derive an expression for the work that must be done by an external agent on the second particle to increase the radius of the circle of motion to r2."

Homework Equations

external work= change in U= Change in KE
U= QqK/r

The Attempt at a Solution

I assumed that the speed at r2 was the same, which might be a wrong assumption and then said the change in U= work. The work will also be negative since the force points in the opposite direction of the displacement. This works out to be w=Ur1-Ur2= kqQ(1/r1-1/r2). However the answer in the book says w=(kqQ/2)(1/r1-1/r2). I'm very confused where the 1/2 has come from. I think maybe there is a relationship between the initial and final kinetic energy that I'm not getting?

thank you

 

[Astronuc]

I assumed that the speed at r2 was the same,

If the particle is in a stable orbit, then the centripetal force at r2 must balance the attractive force, so what must happen to the tangential speed?

 

[supersunshine]

Hi,
thanks for the tip. I set Fcent=Felec= mv^2/r=kQq/r...solving for v^2 I got V1^2= kQq/mr1 and V2^2=kQq/mr2. I pluged these values into the equation KE1+U1=KE2+U2= Wext and got wext= 3KQq/2 (r2-r1/r1r2). Where am I going wrong?

 

[mukundpa]

Fcent=Felec= mv^2/r=kQq/r...

I think
Felec = kQq/r^2

 

[mukundpa]

That you have corrected but what about the electrostatic potential energy? Is it positive, or negative!

 

[supersunshine]

thanks, i added change in KE+ change in PE and got 3/2, but i think i should have subtracted KE-PE and 1/2 kQq(1/r1-1/r2)