# Potential Difference and Potential Near a Charged Sheet

1. Oct 9, 2007

1. The problem statement, all variables and given/known data
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

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

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

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

3. 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?

2. Oct 10, 2007

### learningphysics

Looks right to me.