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Electric Potential and Kinematics

  1. Apr 22, 2006 #1
    Hi, I'm currently doing a practice problem and I need help in solving it.

    Ok first the problem:

    a)Ok, the potential energy of the electron is:

    [tex]U = -eV_R[/tex]

    To find V of R we need to refer to electric field:

    [tex]V_f - V_i = -\int_{i}^{f} \vec{E} \cdot d\vec{R}[/tex]
    [tex]V(R) - V(\infty) = -\frac{Q}{4\pi\varepsilon_\varnothing} \int_{R}^{\infty} \frac{dR}{R}[/tex]
    [tex]V_R = \frac{1}{4\pi\varepsilon_\varnothing}\frac{Q}{R}[/tex]


    [tex]U = -eV_R[/tex]
    [tex]U = \frac{1}{4\pi\varepsilon_\varnothing}\frac{-eQ}{R}[/tex]

    B)This is the one I'm having trouble with. I don't know what the final potential energy is or out to get it.

    Here's what I have so far:

    [tex]U_\varnothing = K_f + U_f[/tex]
    [tex]\frac{1}{4\pi\varepsilon_\varnothing}\frac{-eQ}{R} = \frac{1}{2}mv^2 + U_f[/tex]

    So what do I do for U of f?
    Last edited: Apr 22, 2006
  2. jcsd
  3. Apr 22, 2006 #2
    You'll first have to find the electric field inside the sphere using Gauss's law. Then integrate that from 0 to R to get the potential at the center, designating the potential at the surface zero.
  4. Apr 22, 2006 #3
    That means:

    [tex]\int \vec{E} \cdot d\vec{S} = \frac{Q_{enclosed}}{\varepsilon_\varnothing}[/tex]
    [tex]\int E \cdot dS cos(0) = \frac{Q_{enclosed}}{\varepsilon_\varnothing}[/tex]
    [tex]E \cdot 4 \pi R^2 = \frac{Q_{enclosed}}{\varepsilon_\varnothing}[/tex]

    Charge is intact, so...
    [tex]\int E = \frac{1}{4\pi\varepsilon_\varnothing} \frac{|Q|}{R^2}[/tex]

    Which is the same for pointlike charges...


    [tex]V_f - V_i = -\int_{i}^{f} \vec{E} \cdot d\vec{R}[/tex]
    [tex]V(R) - V(0) = -\int_{0}^{R} \vec{E} \cdot d\vec{R}[/tex]
    [tex]V(R) - V(0) = -\frac{Q}{4\pi\varepsilon_\varnothing} \int_{0}^{R} \frac{dR}{R}[/tex]

    eh there's a division by zero. Does this mean that there's no potential energy in the center? That's what I assumed at the begining, but now I'm lost. :confused:
    Last edited: Apr 23, 2006
  5. Apr 23, 2006 #4
    I think you've forgotten Gauss' Law: :wink:
    [tex]\int \vec{E} \cdot d\vec{A} = \frac{q_{enc}}{\epsilon_0}[/tex]
  6. Apr 23, 2006 #5
    No I haven't. it's in my previous post. Am I complicating things?
  7. Apr 23, 2006 #6
    You're trying to find the electric field at any distance [tex]r[/tex] from the center of the sphere. So:
    [tex]E \cdot 4 \pi r^2 = \frac{\rho V}{\epsilon_0}[/tex]
    [tex]\rho = \frac{Q}{\frac{4}{3} \pi R^3}[/tex]
    [tex]V = \frac{4}{3} \pi r^3[/tex]
    (Notice r is not the same as R).
    Last edited: Apr 23, 2006
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