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Potential Difference and Potential Near a Charged Sheet

by Badger
Tags: charged, difference, potential, sheet
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Oct9-07, 10:15 PM
P: 29
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

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?
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Oct10-07, 11:14 AM
HW Helper
P: 4,124
Looks right to me.

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