Calculation of the amount of work done using potentials

[Physicslearner500039]

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 constant speed in a circle of radius 1'10 centered at P. Derive an expression for the work W that must be done by an external agent on the second particle to increase the radius of the circle of motion to r2

Relevant Equations

NA

This is the diagram in mind

Some of my assumptions like the particle q will be moving with some initial velocity V1 in the circle of radius of r1 and when it moves to the circle of radius r2 its velocity will be V2 ( > V1) since its potential energy is decreased.
The amount of work done by external agent is change in potential energy
Uf - Ui = W (Work done by external agent by definition).

The initial configuration is rotating in circle of radius r1.
The potential energy in this configuration is Ui = Q*q/(4*π*ε0*r1);
The kinetic energy is Ki = 0.5*m*V1^2;
Ei = Ui+Ki

The total energy in cfg2 is
Ef = Q*q/(4*π*ε0*r2) + 0.5*m*V2^2;

But there are so many unknowns, I am not sure how to calculate the work done. Please advise.

 

[etotheipi]

There are a few things that are wrong here. Firstly, your diagram should have two concentric circles, since the outermost orbit that you have drawn is not possible in its current state.

when it moves to the circle of radius r2 its velocity will be V2 ( > V1) since its potential energy is decreased.

Secondly, you quote $U = \frac{Q_1 Q_2}{4\pi\epsilon_{0} r}$. Here, $Q_1 = Q$ and $Q_2 = -q$, so $U$ is negative (indicative of a bound state). Moving the charge $-q$ to an orbit of larger radius should increase the potential energy of the system. This increase is easy to compute from the definitions.

You need to think a little more about how you get the kinetic energy of the two orbits. The standard approach is to apply the circular motion condition. A trick would be to use the virial theorem.

Once you've computed the total energies of both orbits separately ($E = T + U$), it should be straightforward to find the external work needed.

 

[haruspex]

will be V2 ( > V1) since its potential energy is decreased.

The change in speed is not to do with the change in PE. Consider each radius as constituting a stable orbit, with the attraction providing the centripetal force.

 

[Physicslearner500039]

I think I made some progress but not completely. The force of attraction between the charges should be equal to the centripetal force
$F = q*Q/(4πε_0r_1^2)$
This is equal to the centripetal force
$m*V^2/r_1 = (q*Q)/(4πε_0r_1^2); \\ 0.5*m*V^2 = (q*Q)/(8πε_0r_1); \\ KEi = (q*Q)/(8πε_0r_1); \\ Ui = (-q*Q)/(4πε_0r_1); \\ KE_f = (q*Q)/(8πε_0r_2); \\ U_f = (-q*Q)/(4πε_0r_2); \\$
From the book i read the work done is only the change in potential energy, so how do i include the kinetic energy in this equation? Please advise.

 

[etotheipi]

From the book i read the work done is only the change in potential energy, so how do i include the kinetic energy in this equation? Please advise.

This isn't quite right, though it's often not too clearly stated.

The change in potential energy associated with a certain conservative force equals by definition the negative of the work done by that conservative force; this just comes from $\vec{F} = -\nabla U$. This is always true, but is not quite what you need here. I assume this was the definition you read in your book.

Instead, this question is asking you to consider an external force doing external work on the system in order to increase its total energy. For any system, in the absence of other mechanisms of energy transfer (like heat), the total work done on that system by external forces equals the change in its total energy, $E$. Really that just means $W_{ext} = \Delta E = \Delta KE + \Delta U$. This is just conservation of energy, or you can think of it in more of a 'first law of thermodynamics' sense.

 

[Physicslearner500039]

Ok, Thank you. The final answer is
$W = (q*Q)/(8*π*ε_0)(1/r_1 - 1/r_2)$