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Chapter 22: Problem 4

The electric field at the center of a uniformly charged ring is obviously zero, yet Example 22.6 shows that the potential at the center isn't zero. How is this possible?

Short Answer

Expert verified
The electric field at the center of a uniformly charged ring is zero due to the vector nature of the electric field where all the small fields cancel each other out. But the electric potential is not zero due to its scalar nature where all the small potentials from different points of the ring add up to give a non-zero finite value.

Step by step solution

01

Define Electric Field and Potential

The electric field, denoted E, at a point in space is defined as the force that a positive test charge would experience at that point, divided by the charge's magnitude. On the other hand, the electric potential, denoted V, at a point in space is the total work done by the electric field in bringing a unit charge from infinity to that point.
02

Explain Electric Field at the Center of a Charged Ring

In a uniformly charged ring, the electric field at the center is zero because the field is vector in nature. Each small part of the ring contributes a little field, and these fields have different directions. When all these little fields are added as vectors considering their directions, they cancel each other out, making the total electric field at the center zero.
03

Explain Electric Potential at the Center of a Charged Ring

The electric potential, on the other hand, is scalar in nature. It doesn't have a direction, but only magnitude. So, when calculating the potential at the center due to each small part of the ring, we don't need to consider their direction. Therefore, they don't cancel out each other as in the case of the electric field. Instead, they add up to a certain value, making the potential at the center not zero but a finite value depending on the total ring charge and its radius.

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Most popular questions from this chapter

A sphere of radius \(R\) carries a nonuniform but spherically symmetric volume charge density that results in an electric field in the sphere given by \(\vec{E}=E_{0}(r / R)^{2} \hat{r},\) where \(E_{0}\) is a constant. Find the potential difference from the sphere's surface to its center.

Two identical charges \(q\) lie on the \(x\) -axis at \(\pm a\). (a) Find an expression for the potential at all points in the \(x\) -y plane. (b) Show that your result reduces to the potential of a point charge for distances large compared with \(a\).

The classical picture of the hydrogen atom has the electron orbiting 0.0529 nm from the proton. What's the electric potential associated with the proton's electric field at this distance?

A disk of radius \(a\) carries nonuniform surface charge density \(\sigma=\sigma_{0}(r / a),\) where \(\sigma_{0}\) is a constant. (a) Find the potential at an arbitrary point \(x\) on the disk axis, where \(x=0\) is the disk center. (b) Use the result of (a) to find the electric field on the disk axis, and (c) show that the field reduces to an expected form for \(x \gg a\)

Would a free electron move toward higher or lower potential?
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