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I thought the electric potential at point B would be positive because placing a positive charge at point B would cause it to move releasing energy while doing so. The answer key states that the electric potential is 0 at point B and while the math seems to support that because $k\frac Qr + k\frac{-Q}r = 0$, I don't understand conceptually how the potential is 0.

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  • $\begingroup$ It it is at potential zero, it would move toward points at which the potential is negative. So moving doesn’t imply that it is at positive potential. Think about the work done as the particle moves from A at infinity to B. $\endgroup$
    – G. Smith
    Jun 11, 2019 at 20:00
  • $\begingroup$ If we could trace a line from infinity to center of the dipole such that all the field lines are perpendicular to this line then we have that infinity and centre are equipotential ( a hint only ) $\endgroup$ Jun 12, 2019 at 9:11

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Of course, it is 0.

And it is true that, if you place a third charge in that midpoint, that charge will move towards one of the ends, depending on the sign. If the extra charge is possitive, it will move towards the negative, of course.

That is because the electric field is not 0. Charges move according to forces. Well, to be precise, charges can move anyhow, but if you place a charge at rest, it will accelerate according to the force. This is well known: $F=ma$. And the force is proportional to the electric field.

So, the potential does not have such an immediate interpretation. The "meaning" of potential is more subtle. You can interpret that it costs $0J$ to move the charge from infinity to that midpoint. You can do it with no work. That's the possible interpretation of potential.

If you want a more intuitive interpretation of potential, you should recall that

$$F= q\frac{dV}{dr}$$

so the force is proportional to the gradient of potential. That means that you should not care about the potential, but the variation of the potential.

$V$ can be 0 at the midpoint, but you should check if the potential is varying around that point, and it is. IT does vary. That variation if what causes the charge to move.

A good analogy to compare is gravity: height does not matter, you can be stable in any floor of a skyscraper. What matters is the variation of height, that is, the slope. The slope causes things to fall.

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At each point on line $AB$ there are two forces on a positive test charges; a repulsive force due to the positive dipole charge and an attractive force due to the negative dipole charge.
Both forces have the same magnitude and inclined by the same angle relative to the line $AB$.
This means that the net force on the test charge is at right angles to the line $AB$.
To make the test charge move along line $AB$ an external force must be applied at right angles to line $AB$ and that external force does no work moving along line $AB$ as the force is at right angles to its displacement. .
So no work is done by the external force moving the test charge from infinity (the zero of potential) to point $B$ which is therefore at a potential of zero.

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  • $\begingroup$ I was busy editing and didn't see your answer before posting mine. Kinda says the same thing. Sorry about that. $\endgroup$
    – Bob D
    Jun 11, 2019 at 21:37
  • $\begingroup$ @BobD Do not worry and i hope that the OP appreciates the variety of answers! $\endgroup$
    – Farcher
    Jun 11, 2019 at 21:41
  • $\begingroup$ Thanks. I didn't want you to think I was plagiarizing. $\endgroup$
    – Bob D
    Jun 11, 2019 at 21:54
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The potential difference $V$ between two points is the work per unit charge required to move the charge between the points. See the figure below.

What is the work required to bring a charge Q (+or -) from point A an infinite distance away from B to point B (B located between the charges $Q_1$ and $Q_2$)? Answer 0. This is because the force exerted on the charge $Q$ by the electric field between $Q_1$ and $Q_2$ is always perpendicular to the displacement of the charge in going from A to B. That means zero work is done. Thus the potential at point A, as well as at every point in the path of the charge $Q$ going from A to B is zero.

Another way to look at it, which you might find more intuitive, is the farther away you move point A from point B, the closer together the two charges $Q_1$ and $Q_2$ appear. The closer the charges appear, the more the net charge appears to be zero, which means there would no potential at point A an infinite distance from B.

Hope this helps.

enter image description here

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