Potential Difference and Potential Near a Charged Sheet

AI Thread Summary
The discussion focuses on calculating the potential difference between two points near a charged sheet with surface charge density +σ. The electric field generated by the sheet is constant at E = σ/(2ε₀) and directed away from the sheet. For Part A, the potential difference V_AB is derived as V_AB = -E(y_1 - y_2), indicating a linear relationship based on the vertical positions of points A and B. In Part B, when considering the potential at a distance y_1 from the sheet, the potential is evaluated to be infinite, as the reference point at y = ±∞ is set to zero. The calculations confirm that the approach taken is correct.
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Homework Statement


Let {\rm A} = \left(x_1,y_1 \right) and {\rm B} = \left( x_2,y_2 \right) be two points near and on the same side of a charged sheet with surface charge density + \sigma . The electric field \vec{E} due to such a charged sheet has magnitude E = \frac {\sigma}{2 \epsilon_0} everywhere, and the field points away from the sheet, as shown in the diagram. View Figure
184996.jpg


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

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


Homework Equations


\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<br /> = C_x (x_1 - x_2) + C_y(y_1 - y_2)

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



The Attempt at a Solution


Part A.
V_{\rm AB} = V_{\rm A} - V_{\rm B}= \left(-E\right)\left(y_{1}-y_{2}\right)

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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