# Work required to move charged particle traveling in a circle

1. Feb 28, 2012

### xlava

1. The problem statement, all variables and given/known data
Hi everyone and thank you in advance for your time. I just had this problem on a physics exam (that everyone in the class bombed, and I mean everyone, including the best students). I honestly couldn't care less about the grade, but I really want to understand where I went wrong on this one... a picture outlining the problem parameters is attached.

The problem is asking how much work is required to move a particle charge q, mass m traveling at constant velocity v in a circle, from r1 to r2.

2. Relevant equations

Kinetic Energy = $\frac{1}{2}$mv2
Electric Potential Energy = $\frac{kQq}{r}$

3. The attempt at a solution

The particle has initial kinetic and potential energy, hence the total energy of the system when the particle is on the r1 line is:

$\frac{1}{2}$mv02 + $\frac{kQq}{r1}$

and the final energy of the system at r2 will be:

$\frac{1}{2}$mvf2 + $\frac{kQq}{r2}$

So would the work done by an outside force simply be equal to the difference in the total energies of the system? This seems to make sense but I have a feeling that its simpler than this... I am also almost certain that voltage comes into play here, but I'm not quite sure where.

Thanks again.

#### Attached Files:

• ###### physics problem help 1.jpg
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2. Feb 28, 2012

### Staff: Mentor

The difference in total energy for the two configurations should give you the energy required to perform the change, hence the work.

Beware if the charges have the same sign! It means that some external agency is already holding the particle in orbit, and allowing it to move to a larger orbit will do work on that agency rather than vice-verse.

The voltage (potential) at a location due to the central charge is kQ/r. it already shows up in the potential energy term of your total energy. To wedge it into the kinetic energy portion you can equate the centripetal acceleration with the Coulomb force at a given orbit and find a replacement for $v^2$.

3. Feb 29, 2012

### xlava

Yeah I'm an idiot, completely confused myself with voltage equations and stuff.

Its just -kQq/2r (orbital energy), and the difference is what I'm looking for.

Sorry for annoying you guys with this... thanks