Potential Difference and Potential Near a Charged Sheet

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Homework Statement


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
Diagram of a uniformly charged sheet with surface charge density sigma marked by plus signs along the bottom, with upward pointing electric field lines labelled E. Points A(x1, y1) and B(x2, y2) sit above the sheet, joined by a dashed curved path.


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


Homework 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<br /> = 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]



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?