Why is electric potential 0 in this case?

Question

On a test, we had a question where there are 4 point charges at the vertices of a square. The 2 charges at the upper vertices have charges of +q and the 2 charges at the lower vertices have charges of -q. The magnitude of the charges are equal. According to the answer sheet, the electric potential is 0 along a horizontal line halfway between the 2 upper and 2 lower charges. Why is it 0? Shouldn't the test charge be attracted to either to top or the bottom depending on its charge?

+q        +q

------------    <--- 0 electric potential

-q        -q

Answer 1

The result comes from superposition--the idea that the total potential on a test charges comes from finding the potential due to each of the four charges separately and then simply adding them up.

Label each of the charges as follows: TL for top-left, TR for top-right, BL for bottom left, and BR for bottom right.

TL        TR

------------    <--- 0 electric potential

BL        BR

Each point on the line is the exact same distance away from TL as it is from BL. That means that their potential, $V = kq/R$ will have the same value except for opposite signs. This is because their charges $q$ have the same magnitude, and their distance from the line, $R$, is the same. When you add two numbers that have the same value (magnitude) but opposite signs, such as -3 + 3, they add up to zero, and so TL and BL's potentials cancel each other out along the line.

Each point on the line is the exact same distance away from TR as it is from BR. Hence the magnitude of the potential due to TR is the exact same as the magnitude of the potential due to BR. They will, however, have opposite signs, and so they too cancel out.

Answer 2

Simply ask, if we take a charge of either sign from the infinite extension of the line (say from the right) and move it along the line, what force will it feel in the direction of the line?

Clearly, it feels equal attraction and repulsion from the positive and negative charges, and hence the net force along the line is zero.

The basic definition of work done is force times distance.

Therefore the work done in moving along the line is zero.

Therefore the potential difference b/w infinity and any point on the line is zero.

Answer 3

For this to make sense, you'll need to know the relation between work and electric potential. Here's a PDF that does some explaining if this is new to you.

First, draw the electric field near the line of symmetry you've drawn. If you do this correctly, you should find the direction of $\vec{E}$ is all the same along that line. Now imagine trying to move a test charge along the line at constant speed. How much work would you need to do? (Or, some people like to imagine the work done by the electric field when the particle is constrained to move on that path.)

As another answer pointed out, really all this tells you is that the potential along that line is constant; it's a choice to set it to zero.

Answer 4

In a sense, the question is ambiguous because it doesn't tell you what to use as your $0$ of electric potential. But presumably your professor made it clear that in situations like these you are supposed to assume that the potential at infinity defines your zero of potential. With this in place, the question is asking what is the potential difference between infinity and the horizontal line?

The easiest way to see that the potential is zero everywhere on the line is by looking at a transformation that is a symmetry of the problem. The symmetry operation we will look at is inverting the four charges and reflecting them over the horizontal line.

This symmetry leaves the charge distribution the same and keeps infinity at infinity. Thus the potential difference between a point on the line and infinity must be the same after this transformation.

However, we also know that reflections have no effect on potential difference and inverting the charge distribution multiplies the potential difference by $-1$. Therefore the effect of the composition of the two things must be to multiply the potential difference by $-1$.

Thus if $V$ is the potential difference between infinity and a point on the line, and $V_f$ is the potential difference after the transformation, then, since the transformation is a symmetry operation, we must have $V=V_f$, but on the other hand, since the symmetry inverts the charges we must have $V_f=-V$. From this we conclude $V=0$.

Answer 5

Any point lying on the horizontal line passing halfway between the two upper and lower charges is equidistant from both charges. Hence magnitudes of electric field intensity on charge(either positive or negative) placed at any point on this line are equal. When we resolve the electric field vector into its components we see that the components along the line passing halfway between the upper and lower charges cancel each other out. Hence no electrostatic force acts along that line on the charge placed on it. So, when we eject a charge into the field, no force would be required to move the charge along the referred line. Hence no work would be done in moving the charge along that line resulting in 0 potential along the line.

Answer 6

If you take a point in the horizontal line in the middle,then the potential caused by the upper left charge will be the same as the lower left charge but with opposite signs.The same goes for the two charges on the right.Just take a random point on that line and let a be the distance from upper or lower left charges to the point and b the distance from the upper or lower right charges.Then you can see that they annihilate each other!Like an answer said before me,you can use your intuition for the superposition principle to conclude that immediately.Or you can just do the calculations with the a and b that i told you.