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Potential Difference and Potential Near a Charged Sheet

  1. Oct 9, 2007 #1
    1. The problem statement, all variables and given/known data
    Let [tex]{\rm A} = \left(x_1,y_1 \right)[/tex] and [tex]{\rm B} = \left( x_2,y_2 \right)[/tex] be two points near and on the same side of a charged sheet with surface charge density [tex]+ \sigma[/tex] . The electric field [tex]\vec{E}[/tex] due to such a charged sheet has magnitude [tex]E = \frac {\sigma}{2 \epsilon_0}[/tex] everywhere, and the field points away from the sheet, as shown in the diagram. View Figure
    [​IMG]

    Part A
    What is the potential difference [tex]V_{\rm AB} = V_{\rm A} - V_{\rm B}[/tex] between points A and B?

    Part B
    If the potential at [tex]y = \pm \infty[/tex] is taken to be zero, what is the value of the potential at a point [tex]V_A[/tex] at some positive distance [tex]y_1[/tex] from the surface of the sheet?
    choices are:
    1. infinity
    2. negative infinity
    3. 0
    4. -E * y_1


    2. Relevant equations
    [tex]\int_{\rm B}^{\rm A} \vec{C} \cdot d\vec{\ell} = \int_{x_2}^{x_1} C_x\,dx + \int_{y_2}^{y_1} C_y\,dy
    = C_x (x_1 - x_2) + C_y(y_1 - y_2)[/tex]

    [tex]V_{\rm AB}= -\int _B^A \vec{E}\cdot d\vec{l}[/tex]



    3. The attempt at a solution
    Part A.
    [tex] V_{\rm AB} = V_{\rm A} - V_{\rm B}= \left(-E\right)\left(y_{1}-y_{2}\right) [/tex]

    Part B.
    I figure I'd use the equation I got in part A and set the bottom of the E field at y=0.

    In which case
    V = -E (y_1 - infinity) = infinity

    am i on the right track?
     
  2. jcsd
  3. Oct 10, 2007 #2

    learningphysics

    User Avatar
    Homework Helper

    Looks right to me.
     
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