Potential Difference and Potential Near a Charged Sheetby Badger Tags: charged, difference, potential, sheet 

#1
Oct907, 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? 



#2
Oct1007, 11:14 AM

HW Helper
P: 4,125

Looks right to me.



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