# Electric Potential Energy Inside a Charged Sphere

1. Apr 23, 2015

### Yoni V

1. The problem statement, all variables and given/known data
A point charge q<0 lies just outside a uniformly and positively (non-conducting) charged ball.
Assume the charge can pass through the ball freely. Describe the motion of the charge.

2. Relevant equations
Coulomb's force law, energy equation.

3. The attempt at a solution
Obviously this is simple harmonic motion and easily solved using coulomb's force law and Gauss' law, so I wouldn't even bother writing it down.
The only thing that puzzles me is the energy of the system. We get:
U(r)=kq[4/3πr3ρ]/r=kq4/3πρr2

Since q<0, the equation yields negative values inside the sphere and 0 for r=0.

This implies that the motion from the edge of the ball towards the center requires energy, which contradicts the obvious spontaneous motion of the point charge.

What am I misinterpreting?

Thanks

2. Apr 23, 2015

### haruspex

Are you sure it's SHM? What is the field inside the sphere?

3. Apr 23, 2015

### Yoni V

Well, it is equivalent to the gravitational example of drilling a tunnel through the center of the earth. Both are inverse-square forces and can be proved that the effective force is only due to the inner portion of the ball.
Using Coulomb and Gauss laws we get:
F=k(4/3πr3ρ)(-q)/r2=-(4/3πρkq)r=mqa
Which gives SHM with ω2=4/3πρkq/mq

Am I mistaken?

4. Apr 23, 2015

### haruspex

Where does the charge q start?

5. Apr 23, 2015

### Yoni V

At the outer edge of the ball (say at r=R)

6. Apr 23, 2015

### haruspex

It says 'outside'.

7. Apr 23, 2015

### haruspex

Anyway, to answer your original question, potential energy is always relative to an arbitrary reference value. You have chosen it to be zero at the sphere surface. That will make it negative inside. Not sure why you think that means it will require energy to penetrate the sphere. Just think that through again.

8. Apr 23, 2015

### Yoni V

Oh, I meant just on the edge, such that it can "fall" freely into the ball.
English is not mother tongue, so sorry for being unclear.

Anyway, yes- the value itself is meaningless, but nonetheless given the above equation it increases as the value of r decreases.
Using the formula for the potential energy U=kq1q2/r (+ const.) and applying the superposition principle and Gauss' law we get
U(r)=(4/3)πρkqr2 (+ const=0).
Because q<0 and ρ>0 the energy is negative for all r>0 and zero for r=0, which means that the energy increases as I said above.

Surely my derivation of the potential energy is somehow incorrect, or perhaps it's my interpretation, but I don't understand what's wrong with it.

Thanks again

9. Apr 23, 2015

### haruspex

Oh, I'm sorry - it's me that needed to read what you wrote again.
It's not just the energy that has the wrong sign, the force does too. It seems to stem from this:
Why have you inserted a minus sign on the q?