# Potential due to Electromagnetic field

• spark693
In summary: But I don't understand why the field does work.In summary, the change in potential for a proton moving from a point with a potential of +260 V to a point with a potential of -48.0 V is -308 V. To find the work done by the electric field, we multiply this change in potential by the charge of the proton, giving us a negative value in Joules. This means that the electric field is doing negative work, which is the opposite direction of the motion of the proton. However, upon further discussion, it is determined that the field is actually doing positive work in this scenario.
spark693

## Homework Statement

How much work does the electric field do in moving a proton from a point with a potential of +260 V to a point where it is -48.0 V? Express your answer in electron volts.

V = p(A) - p(B)

## The Attempt at a Solution

V= 260 [v] - (-48.0[V]) = 3.08×102 [V]

I got the right answer, but the wrong sign. I'm not sure if this has anything to do with a proton moving from a positive to negative potential (this would only be important if it was moving from a positive to negative charge?).
Why would the answer be negative?

Last edited:
The change in potential is the potential at the final location minus the potential at the initial location. The proton's starting at a location where the potential is +260V and ending up at a location where it's -48.0V. So ΔV = (-48V) - (+260V) = -308 V.

That's the change in potential. But it's not the work done by the field on the electron. To find the work done there's one more step. Do you know what that is? Hint: The Volt is a compound unit which can also be written as Joules/Coulomb.

1 person
To find work, I multiplied by 1.6x10^-19 C which is the charge of one proton, and that gives me work in Joules?

spark693 said:
To find work, I multiplied by 1.6x10^-19 C which is the charge of one proton, and that gives me work in Joules?

Yes indeed. Work = q ΔV

One more question:
What does it mean that the work is negative? I looked up negative work, and the definition given was that it means the force is hindering movement, or that it is opposite to the displacement. I was confused, as I thought that a proton moving from higher potential to lower potential is movement in the direction that the electric field is pointing.

It means that if you wanted to move the proton from +260V do -48V slowly and without letting it accelerate, you would have to apply a continuous force F to keep it from being accelerated That force opposes the direction of motion. The quantity F*d is the work done in moving the charge from A to B. Since the direction of the force opposes the direction of motion, the work is negative.

So the electric field's work is all about not letting the proton accelerate while moving it?
I'm not sure if this is correct, but I'm picturing the proton naturally moving from the point of higher potential to the point of lower potential (requires no work from the electric field), and the electric field works against it (negative work), only so the proton does not accelerate.

I retract my above statements The field will want to accelerate the proton in its move from a location of +260V potential to one of -48.0V. Thus the force due to the field is in the direction of motion, and so the field must be doing positive work in moving the proton from the initial position to the final position. So it looks like the "right answer" that was given to you wasn't right at all, and your own solution is correct after all.

I think I was thinking about the work required by an external agent to move the charge. Here the filed is doing the job.

1 person

## 1. What is potential due to electromagnetic field?

Potential due to electromagnetic field is the energy associated with the presence of an electric field and a magnetic field. It is a measure of the work required to move a unit charge from one point to another in the presence of these fields.

## 2. How is potential due to electromagnetic field calculated?

The potential due to electromagnetic field is calculated by multiplying the electric potential and magnetic potential at a given point. The electric potential is measured in volts and the magnetic potential is measured in teslas.

## 3. What is the difference between scalar and vector potential due to electromagnetic field?

Scalar potential only takes into account the electric potential, while vector potential takes into account both the electric and magnetic potentials. Scalar potential is a scalar quantity, while vector potential is a vector quantity.

## 4. What are some practical applications of potential due to electromagnetic field?

Potential due to electromagnetic field is used in many electronic devices such as computers, smartphones, and televisions. It is also used in power generation and transmission, medical imaging, and communication systems.

## 5. How does potential due to electromagnetic field affect charged particles?

Charged particles will experience a force when placed in an electromagnetic field, which can cause them to move and change their potential energy. The direction and magnitude of this force depend on the strength and direction of the electric and magnetic fields.

• Introductory Physics Homework Help
Replies
6
Views
155
• Introductory Physics Homework Help
Replies
22
Views
1K
• Introductory Physics Homework Help
Replies
1
Views
1K
• Introductory Physics Homework Help
Replies
1
Views
1K
• Introductory Physics Homework Help
Replies
3
Views
526
• Introductory Physics Homework Help
Replies
1
Views
886
• Introductory Physics Homework Help
Replies
11
Views
1K
• Introductory Physics Homework Help
Replies
2
Views
362
• Introductory Physics Homework Help
Replies
4
Views
3K
• Introductory Physics Homework Help
Replies
2
Views
182