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How do you find the electric potential from force?

  1. Feb 11, 2013 #1
    1. The problem statement, all variables and given/known data

    Suppose Coulomb's force is actually

    F= (y)(q1q2)(r hat)/(r^4)

    with y being a constant. Find the electric potential function V(x,y) for a charge Q located at the point x=a, y=b

    2. Relevant equations

    Fq=E

    V=integral (E)


    3. The attempt at a solution

    I am very confused by this problem and do not know if my solution is correct:

    1)

    find electric field using fq=E,

    E=(y)(q1)(r hat) / (r^4)

    2)

    Integrate to find V. Somehow change q1 to Q

    V= integral (E dot Dr)
    V= Qy integral (1/r^4)

    v= -Qy/r^3,

    where r = sqrt((x-A)^3 + (y-B)^2)

    giving

    V (x,y) = - yQ / 3 ((x-a)^2 + (y-b)^2))^(3/2)) + 0


    Is this correct? How does q1 change into Q?
     
  2. jcsd
  3. Feb 11, 2013 #2

    SammyS

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    Not correct.

    Are you sure that's y, and not γ (gamma) ? Whatever it is, it's a constant.

    I'm assuming that [itex]\displaystyle \ \ \vec{F}=\gamma\,\frac{q_{1}q_{1}}{r^4} \hat{r}\ \ [/itex] gives the force exerted on q2 by q1 .

    Then what you for the E field due to q1 makes sense. To find the E field due to Q, simply replace q1 with Q .


    To get the potential, simply integrate.

    [itex]\displaystyle V(r)=-\int \vec{E}(r)\cdot d\vec{r}[/itex]
    [itex]\displaystyle =-\int\gamma\,Qr^{-4}\hat{r}\cdot d\vec{r}[/itex]​
     
  4. Feb 11, 2013 #3
    Thank you.



    I know that dr = dr ir + dθ ir and that dr = dx ix + dy iy.


    How do I know in which situation to use each?
     
  5. Feb 11, 2013 #4

    SammyS

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    Well, [itex]\hat{r}\ \text{ is parallel to }\ d\vec{r}\ \ [/itex] so that [itex]\hat{r}\cdot d\vec{r}=dr\ .[/itex]

    That results in [itex]\displaystyle \int\vec{E}\cdot d\vec{r}=\int E\,dr\ .[/itex]
     
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