# Physics Lesson 16.11.3 - A New Approach on Electric Potential

Welcome to our Physics lesson on A New Approach on Electric Potential, this is the third lesson of our suite of physics lessons covering the topic of Induced Electric Fields, you can find links to the other lessons within this tutorial and access additional physics learning resources below this lesson.

## A New Approach on Electric Potential

As stated in the previous paragraph, both the static and dynamic charges produce electric fields around them. However, they have a fundamental difference: the electric field lines of static charges are open (start from positive charges and end at negative ones) while dynamic charges form closed lines due to their circular path around the loop. Since the electric field produced by dynamic charges is uniform, it is meaningless to talk about electric potential (and potential difference) caused by moving charges.

Therefore, the main difference between electric fields produced by static and dynamic charges is as follows:

"Electric potential has meaning only for static charges, not for dynamic ones."

Let's prove the correctness of the above statement by writing the integral form of the potential difference produced by static charges:

∆V = E ds

When moving in a closed loop at the same distance from the centre of field (in equipotential surfaces), the above expression becomes

∆V = E ds = 0

However, when the magnetic flux changes, we have (from the new version of Faraday's Law discussed earlier in this tutorial)

E ds = -M/dt

This means the integral is not zero, so ΔV ≠ 0. These two different results for the same thing represent a contradiction. Therefore, it is meaningless to discuss about induced electric potential.

### Example 2

How many seconds are needed to change the electric flux from 0.3 Wb to 1.5 Wb if a test charge inserted inside a uniform magnetic field is moving according the circular path shown in the figure? The electric field induced during one complete rotation in the loop is 12 V/m and the loop's diameter is 2 cm.

### Solution 2

Clues:

Φ1 = 0.3 Wb
Φ2 = 1.5 Wb
E = 12 V/m
d = 2 cm = 0.02 m
dt = ?

First, we find the radius of loop. We have

r = d/2
= 0.02 m/2
= 0.01 m

Since the magnetic field is uniform, we can write the integral version of Faraday's Law

E ds = -M/dt

as

E ∙ 2πr = -M/dt

Ignoring the negative sign in the right side of the equation (it just shows direction), we obtain

dt = M/E ∙ 2πr
= Φ21/E ∙ 2πr
= 1.5 Wb-0.3 Wb/(12 V/m) ∙ 2 ∙ 3.14 ∙ (0.01 m)
= 1.59 s

You have reached the end of Physics lesson 16.11.3 A New Approach on Electric Potential. There are 3 lessons in this physics tutorial covering Induced Electric Fields, you can access all the lessons from this tutorial below.

## More Induced Electric Fields Lessons and Learning Resources

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16.11Induced Electric Fields
Lesson IDPhysics Lesson TitleLessonVideo
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16.11.1Induced Electric Fields
16.11.3A New Approach on Electric Potential

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