Electric Potential Energy Inside a Charged Sphere

[Yoni V]

Homework Statement

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.

Homework Equations

Coulomb's force law, energy equation.

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

 

[haruspex]

Assume the charge can pass through the ball freely. Describe the motion of the charge.
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.
Thanks

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

 

[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=m_qa
Which gives SHM with ω²=4/3πρkq/m_q

Am I mistaken?

 

[haruspex]

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=m_qa
Which gives SHM with ω²=4/3πρkq/m_q

Am I mistaken?

Where does the charge q start?

 

[Yoni V]

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

 

[haruspex]

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

It says 'outside'.

 

[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.

 

[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

 

[haruspex]

Because q<0 and ρ>0 the energy is negative for all r>0 and zero for r=0

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:

F=k(4/3)πr³ρ(-q)/r²

Why have you inserted a minus sign on the q?