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Question about equipotential lines and electric fields

  1. Sep 6, 2016 #1
    1. The problem statement, all variables and given/known data
    Hi, I have a question about my lab report that I am doing. The only question that I am having problems to understand is the first one, it says, "why are the equipotential lines near conductor surfaces parallel to the surface and why are they perpendicular to the insulator surface mapped?" I am including a picture of my lab. Now I am not sure if all of it is correct, but the equipotential lines are correct (the equipotential are for a 0 Volt reading and the power source was set to 4 volts, so it is a potential difference I assume. The table where the paper was placed also had resistors, but I am not sure what those are for since we haven't covered it yet in lecture). So I made an explanation about my question in the picture, but I am not sure if it is right. help!!

    2. Relevant equations


    3. The attempt at a solution
    001.jpg
    my solution is on the image
    https://postimg.org/image/yf66y9sqn/ [Broken]
     
    Last edited by a moderator: May 8, 2017
  2. jcsd
  3. Sep 6, 2016 #2

    kuruman

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    Question 1
    Why are the equipotential lines near conductor surfaces parallel to the surface?

    Think of gravitational equipotential surfaces as analogues for electrical equipotential surfaces. What do the gravitational equipotential surfaces near the surface of the Earth look like? Why do you think that is? Carry the same thinking to a a point near the surface of a conductor with curvature.

    Question 2
    Why are they perpendicular to the insulator surface mapped?

    I am not sure what is meant by "insulator surface mapped." It looks like it means the paper on which you mapped your lines. However, your figure shows a circular insulator like a piece of plastic or something. Was such a thing on the paper when you did the experiment?

    In any case, the basic difference between conductors and insulators is that electric field lines cannot exist inside a conductor while they do go through an insulator. Add to this the idea that equipotential surfaces are locally perpendicular to electric field lines.
     
  4. Sep 6, 2016 #3
    well gravity is stronger in the ground than 1000 ft above the ground, so the gravitational field is stronger on the surface of the earth. Is like looking at a hole that gets smaller and smaller as it gets deeper and deeper right? so it converges. The conductor in my graph is a circle of equal radius everywhere, then at any point on the charged outer surface of the circle, it will always have an equal curvature. I am still unsure what it means by "parallel to the surface"?. The surface of the equipotential line follows the circular path of the conducting circle, so it is parallel? So if the equipotential line that touches the circle is constant as well as the electric field, then would the derivative be 0, or does it has something to do with the gradiant?

    For question 2 it was a circle that had wood as insulator.
     
  5. Sep 7, 2016 #4
    i think i understand now. The equipotential lines will get closer to the surface as the voltage gets increased. I only used 4 volts, but maybe a 6 or 10 would get the equipotential lines closer and closer to the surface of the conductor, thus beign parallel to the surface. In the case of the insulator, it will not get charged because insulators dont get charged or voltage. thus, the equipotential lines will go divergent around its surface and perependicular, and convergent to the surface of the conductor. those that makes sense?
     
  6. Sep 7, 2016 #5

    kuruman

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    Electric field lines are perpendicular to the surface of a conductor because if they were not, there would be a component of the electric field parallel to the surface. This assumed component will exert a force on the electrons that are free in the conductor and cause them to move. They will keep moving until they reach a point on the surface where they no longer experience a force parallel to the surface. This is another way of saying that the free electrons in a conductor rearrange themselves in the presence of an external electric field so that the field has only a normal component to the surface. If the electric field lines are perpendicular anywhere on the conductor's surface, and electric field lines are also perpendicular to equipotentials what does that make the equipotentials near the surface?

    Think of the Earth and its gravitational field. Near the surface of the Earth (where we are) the Earth looks flat. The gravitational potential energy is U = mgh (where h is the height above ground and h is much less than the radius of the Earth.) The gravitational potential is energy per unit mass, i.e. V = gh. The gravitational equipotentials are planes parallel to the surface of the Earth at any height h. Carry this thinking to a point at distance h near a charged conductor where h is much less than the radius of curvature of the conductor.

    It is better to say that as the voltage gets increased, the distance between equipotentials differing by the same voltage decreases. That's because the electric field (ΔV / Δx) increases, i.e. you have less Δx for the same ΔV; for example the distance in cm between the 2 V equipotential and the 4 V equipotential will be reduced. This also means that if your conductor is at 10 V and the distance from the surface to the 9 V equipotential (ΔV = 1 V) is, say, 1 cm, when you increase the conductor's potential to 12 V, the distance to the 11 V equipotential (also ΔV = 1 V) will be less than 1 cm.
     
  7. Sep 7, 2016 #6
    if both the surface of the conductor and the equipotential line are perpendicular to the electric field, then it means that since they will be at 90 degrees, then the total work will be zero (Fdcos90=0). so the voltage will stay the same on the surface and on the equipotential line because it takes work to make a change in voltage, and since no change is done, then it is 0. so equipotential lines and the surface are parallel because they do not get a change of voltage. right? and for the insulator force, since no equipotential line can go near it, and the electric fields go inside of it, then the equipotential lines will not have any component near the surface of the insulator and thus it will be perpendicular to the surface. I understood what you mean about the earth gravitational field and the voltage differences, but i am not how to use them to explain my question!!?
     
  8. Sep 8, 2016 #7

    kuruman

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    I mentioned gravitational equipotentials to help you understand electric equipotentials. You seem to have now a basic understanding of what equipotentials are. As the word implies, they are lines or surfaces (more generally) of equal potential. This also means that the potential difference (a.k.a. voltage) between any two points lying on an equipotential is zero which in turns implies that the electrical forces do zero work on charges that are moved from one of these points to the other. Note that the path that the charge follows does not matter; as long as you start and end at the same equipotential, the work done by the electrical forces will be zero. That's a powerful idea that can help you solve a large category of problems in electrostatics.

    Your thinking about the insulator seems to be unclear. You write about "the insulator force". What force is this? Do you mean the force on the insulator? If so. what entity exerts this force? You also seem to believe that equipotentials cannot go near an insulator. I don't know what you mean by "go near". As long as there are electric field lines in space, there will be equipotential surfaces perpendicular to these lines at each point in space. Since an insulator has electric field lines actually going through it, equipotentials can be arbitrarily close to an insulator and in fact inside the insulator. Similarly for a conductor, except that there is no electric field inside the conductor which makes the entire space occupied by the conductor an equipotential.

    I have appended pictures of a conducting and an insulating sphere placed in a uniform electric field. Study them carefully and imagine the equipotential lines everywhere in the space surrounding the spheres in each case. Also note that (a) there is no field inside the conductor while a field weaker than the external field exists in the insulator; (b) the E-field lines are perpendicular to the surface of the conductor while they are not in the case of the insulator. I already explained to you why electric field lines must be perpendicular to the surface of a conductor. Why they are not in the case of an insulator is a topic that is clarified in an intermediate physics course on Electricity and Magnetism and you don't have to worry about it for your lab write up.
     

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  9. Sep 8, 2016 #8
    aaaaa i see to understand now. i understand the whole concept of why equipotential lines most be perpendicular to the electric field now, and I spend some hours online reading about them. My book does not seem to be good at explaining this. So the reason why equipotential lines are parallel to the surface is because since there are eletric fields touching the surface, then the equipotential lines will be parallel to those electric fields and parallel to the conductor surface. (by looking at the picture you attached). And the reason why equipotential lines are perpendicular to the surface of the insulator is because since there will be electric field lines inside of it, then the equipotential lines will be parallel to those electric fields, and thus perpendicular to the surface because the insulator has a constant curvature at every point. All is clear to me now!! thanks!
     
  10. Sep 8, 2016 #9

    kuruman

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    You are welcome. A picture is worth 1000 words. :smile:
     
  11. Sep 8, 2016 #10
    jaja i know!! thanks!!
     
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