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Does equal electric field imply equal potential?

  1. Apr 11, 2013 #1
    1. The problem statement, all variables and given/known data
    Given the E field
    E = 18/R2 R, R is the radial direction.
    Find the electric potential between A and B where A is at +2m and B at -4m, both on the z axis.


    3. The attempt at a solution
    My question is, since E field depends only on R, the distance between the point and the origin, then E field is the same for point B at -4m and point C at +4m. Therefore they are equipotential. Therefore I can compute the potential difference between A at +2m and C at +4m. Is this reasoning correct?
     
  2. jcsd
  3. Apr 11, 2013 #2

    cepheid

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    No, equal electric field does not imply equal electric potential. The electric field is the gradient of the electric potential: E = -∇V, or in one dimension, E = -dV/dr.

    So, if the electric field strength and direction are the same at two points, it doesn't mean that V is the same at those two points, it just means that the derivative of V is the same at those two points.

    Example: electric field between two large parallel plates of uniform surface charge, is the same everywhere between the two plates, but electric potential increases linearly as you move from one plate to the other.

    (The approximately uniform gravitational field close to the surface of the Earth is a totally analogous situation. The gravitational field strength is the same at two points at two different heights, but the gravitational potential is definitely different between those two points.)
     
  4. Apr 11, 2013 #3

    cepheid

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    Hmm...what I said above was the most general answer. HOWEVER in this case, it seems like the problem posesses some symmetry. If dV/dr depends only on r, then V will depend only on r as well. So V is the same at all points that are equidistant from the origin.
     
  5. Apr 11, 2013 #4
    Some questions...
    I'm trying to compute V between two general points for the given E field, just to see what it will give and also for the sake of it.
    We know that

    Vb-Va = [itex]-\int_{A}^{B}\overrightarrow{E}\cdot d\overrightarrow{l}[/itex]

    any idea how I could express [itex]d\overrightarrow{l}[/itex]...?
    also, should [itex]d\overrightarrow{l}[/itex] be from A to B or from B to A?
    is Vb-Va called the potential difference between A and B or the potential difference between B and A?
     
  6. Apr 11, 2013 #5
    I found that [itex]d\overrightarrow{l}=dR\overrightarrow{R}+Rd\phi \overrightarrow{\phi}+dz\overrightarrow{z}[/itex]

    so
    [itex]V_{B}-V_{A}-\int_{A}^{B}\overrightarrow{E}\cdot d\overrightarrow{l}=-\int_{R_{A}}^{R_{B}}\frac{18}{R^{2}}dR=\frac{18}{R_{B}}-\frac{18}{R_{A}}[/itex]
    is that correct? in that case it seems my initial assumption that -4 and 4 are equipotential is correct since it depends only on R which is 4 in both cases.
    is that the potential change when you go from B to A or is it the potential change when you go from A to B?
     
  7. Apr 11, 2013 #6

    cepheid

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    Obviously V_B - V_A is the potential change when you go from A to B. You're taking the difference between the final value and the initial value to compute the change.
     
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