# How can the electric potential at the center of a ring be large if the electric field isn't, and vice versa?

If we have a uniformly charged ring, then at the center of that ring the electric field is zero because each half cancels with the other. However, the electric potential at the center is twice that of one half. Why is that?

My understanding of electric potential is that it is a measurement of how much a charge want to go along the field lines. And since there are no field lines, I'd expected the potential to be zero, since the sum of forces acting on it is zero.

What's even more confusing, is that the potential at the center of a ring with half of it having a positive charge, and the other having a negative charge, is zero, even though the sum of the forces, and thus the electric field, nonzero.

• just to add sth informal: I don't know how the notion "electric potential = measurement how much charges want to go along field lines" can be given any useful interpretation, especially as there is a "gauge freedom" to shift the electric potential by any constant Oct 5, 2016 at 21:53

The relation between the potential $\Phi$ and the electric field $\vec{E}$ is: $$\vec{E} = - \nabla \Phi$$ So the field is not strong when the value of the potential is high but when the local change in the potential is high. Vice versa, if the electric field is zero, the potential might still be at a very high value - it just does not change at that point.
It is a classical confusion for most people learning electrodynamics, but e.g. a potential of zero does not mean that the field there vanishes and vice versa a field of zero does not imply anything about the value of the potential.

Actually in differential form , $E=-\dfrac{dv}{dr}$. So when electric field at centre of ring is zero, $\dfrac{dv}{dr}=0$, i.e, potential is a constant and has non zero value whose expression we obtain.

Another approach would be admiring the fact that potential is a scalar quantity and electric field is a vector quantity. So in case of calculating resultant potential, we only need to see the charges are like or dislike and calculate the resultant algebriacally. So in the first case, the charges on either side of ring are like and so the resultant is algebraically obtained and has a non zero value. But in case of electric field, the resultant is calculated vectorially, i.e, direction comes into play. So, the electric field due to either side of ring, is equal in magnitude but since the direction is antiparellel, the resultant is zero. In second case, the charges on either side are dislike and are equal in magnitude but since the sign is opposite , resultant potential at centre of ring is zero. But, electric field has same direction- away from the positively charged side and towards negatively charged side- and so, the resultant has a non zero value.

Electrical potential by itself is not a measure how much a charge wants to go along the field lines. You can have a point with very high potential and electric field zero, if the neighboring points have the same potential. The electrical potential is only is a measure of the energy you need to move a positive charge from a point of zero potential to the point of this given electrical potential. For the local force on a charge, corresponding to the electric field, the negative gradient of the potential at that location is essential. Therefore, you can have a location with zero potential where you have a nonzero electric field. In addition the electrical potential is not unique. You can chose the zero arbitrarily.

Electric field is a vector, it has a direction. Equal E-fields with opposite direction cancel. Potential is a scalar, it has no direction. Equal potentials with opposite sign cancel.

My understanding of electric potential is that it is a measurement of how much a charge want to go along the field lines.

I think you have gotten this the wrong way around. The electric field is a measure of how much a charge wants to go across the equipotential lines.

What's even more confusing, is that the potential at the center of a ring with half of it having a positive charge, and the other having a negative charge, is zero, even though the sum of the forces, and thus the electric field, is greatest at the center.

The potential is zero at the centre, but the electric field is not greatest at the centre. It is greatest close to the charges, and least in the centre.

You have some incorrect information. You need to go back to your textbook to get the correct information, then it might not be so confusing.

• I understand the math. Its the physical explanation that am fuzzy on Oct 5, 2016 at 21:17

Electric potential is defined as the work done in taking a unit charge from infinity to that point.

Outside the centre of the ring there is a field everywhere so you have to do work to get the charge from infinity through that field and then all the way to the centre. The fact that electric field is zero at the centre does not change the fact that we have to take the charge from infinity to that point all the way through path having electric field.

Thus electric potential is not zero at the centre.

After reading the answers here and solving on my own, I come to a conclusion that V(electric potential) has a constant value at the center of a uniformly charged ring.

• The potential is constant at every point in space if the charges are not moving. This is not unique for the center. What do you mean by "constant"? Jul 16, 2018 at 19:17
• I just know that, at the center of any unformly charged ring, electric poential, V=charge per unit length/(2× permitivity of the free space) Jul 17, 2018 at 10:43
• Am I correct or not...? Jul 17, 2018 at 10:44
• @AaronStevens I know I am wrong at "constant"... :D Jul 17, 2018 at 10:46