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Homework Help: Electric field as a function - potential

  1. Jun 9, 2008 #1
    electric field as a function -- potential

    1. The problem statement, all variables and given/known data

    assume that the electric field in space is given by E = E_o*e^(-r/R) where r is the radial distance away from the origin and E_o and R are constants. E points away fro the origin. Calculate the electric potential at any point r if zero potential is taken at r = +infinity.

    i should get electric potential V(r) = E_o*R*e^(-r/R)
    2. Relevant equations

    point charge electric potential V = V(r) = kq/r where k is constant = 9*10^9, q is charge, r is distance

    electric potential difference deltaV_AB = V_B =V_A = - [<integral>E*dr] from r_A to r_B

    3. The attempt at a solution

    what does it mean that E points away from the origin, how does knowing that affect the problem?

    what is the integral of the E function, how integrate the e^(-r/R) portion specifically?

    if i let r = infinity, then in the E function then e^(-infinity) = 0, so the function goes to zero

    any tips on how to get started appreciated...
  2. jcsd
  3. Jun 9, 2008 #2


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    Homework Helper


    The integral is:

    \Delta V_{ab} = -\int\limits_{a}^{b} \vec E\cdot d\vec r

    and so since there is a dot product inside the integral, you need to know the (relative) directions of [itex]d\vec r[/itex] and [itex]\vec E[/itex] in order to write down the integral for this particular problem.
  4. Jun 9, 2008 #3
    the problem states that E points away from the origin. and r is the radial distance away from the origin. so does that make E negative and dr positive?

    so now do i sub in the function in for E in the integral, how do i integrate e^(-r/R)?

    since i want to calculate for electric potential V, is the charge at the origin, thus a point charge?
  5. Jun 9, 2008 #4


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    Science Advisor

    No, it means that r and E both point in the same direction- so their dot product is just the product of their lengths.

    Use the substitution u= r/R.

    ??There is no mention of a "charge at the origin", just an electric force field- no mention of what causes the force field.
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