0
$\begingroup$

We are taking the electrostatic course in physics class, and I was wondering about some things related to potential, potential energy, and electric field. Imagine two identical particles with opposite charges separated by a distance $d$. In the middle of the two charges, half of $d$, there a test charge is placed. The test charge starts to move to the right, assuming the negative is on the right and the positive is on the left. But if we were to calculate the potential at that point, it would be zero which also implies no energy. The particle, however, accelerates to the right to the negative charge with some energy. Where did that energy come from? You might respond to the question simply by saying, the energy came from the electric field because the motion is caused due to the electric field (or the force of attraction and repulsion contributed by each charge). Then, what do we mean by zero potential/potential energy? What energy are we talking about when we say potential energy or and the kinetic energy of the particle caused by the electric field? Isn't the potential energy defined by the same particles as the electric field or force?

$\endgroup$
  • $\begingroup$ I think you are missing a critical point here. If the test particle starts to move from a stationary position, then there must be a force acting on it. In the case of two stationary, oppositely charged particles, the electric field between them is NOT zero since $\mathbf{E} = -\nabla \phi \neq 0$ at all points between the particles. $\endgroup$ – honeste_vivere Jan 24 '18 at 21:26
  • $\begingroup$ @sammygerbil - But I know the electric field is not zero ( the derivative of the negative of the potential is not zero). That is the question asked there. My question is what is the difference between the energy of the potential and the energy caused by the electric field to move the test charge. $\endgroup$ – ASB Jan 24 '18 at 21:44
1
$\begingroup$

Saying that the potential energy is zero at that point does not make much sense.

Only potential energy differences matter.

  • Think of a ball on the floor. The gravitational potential energy is zero. But only because the floor is our reference. If there is a hole in the floor, then the ball will roll down there, because that position has lower gravitational potential.

The gravitational potential energy in the hole is negative. That does not make much sense in itself. It tells us nothing. Only the difference compared to some other point matters.

  • The ball would not roll to some other point at the same floor because the potentials are the same. And a ball on a shelf would also not roll to another point on that shelf, even though both of those points have gravitational potential energy, because the difference is zero.

Similar for electric potentials. If you calculate a zero electric potential at the starting point of the charge, then it is because you use that same point as your reference. That alone tells you nothing. But the charge sees that the potential gets lower, if it moves closer to the source. So it starts moving. The potential becomes negative seen from the same reference. That doesn't say anything in itself. Only the difference is important.

Bottom line: Whenever there is a potential energy difference, there is a force trying to make the object move. Regardless of the actual potential energy values.

$\endgroup$
1
$\begingroup$

Two things to point out at first:

(a) The value of the potential at a point, P, has, in itself, no bearing on whether a charge initially at rest will move away from that point. What is relevant is the difference in potential between that point and neighbouring points. A positive 'test' charge at P will experience a force in the direction of steepest drop in potential relative to P. Mathematically, the electric field strength is related to the potential gradient by$$\vec{E}=-\vec {\text{grad}}\ (V)$$

(b) It would make no difference to the Physics if we added or subtracted the same constant value to the potentials of every point. So there's little significance in the potential being zero midway between the source charges. It's only differences in potential that matter.

The kinetic energy of the test charge does indeed come from the electric field. As the charge moves away from this midpoint it loses potential energy and gains kinetic energy. Equivalently, work is done on it by the force it experiences from the electric field.

$\endgroup$
1
$\begingroup$

The potential is the potential energy of the test charge divided by the test charge. You could also think of the test charge having the charge 1. In your example of two charges with opposite sign, you are using Coulomb's Law for the potential (or potential energy) of a test charge. You also assume that the potential is zero at infinity, so that at the normal plane going through the midpoint of the line connecting the charges the potential is also zero. You have to keep in mind that the potential and potential energy is only defined up to an arbitrary constant because the electric field ( and thus force on the test charge) is determined by the negative gradient of the potential and energetically only differences of potential are important. To chose the potential zero at infinity is only one choice of the arbitrary constant, you could add an arbitrary constant to the electrostatic Coulomb potentials. Then the potential (and potential energy of a test charge) at the midplane would not be zero but that constant.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.