hw5 - Physics 21 Fall 2009 Solution to HW-5 Potential near...

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Physics 21 Fall, 2009 Solution to HW-5 Potential near a Charged Sheet Let A = ( x 1 , y 1 ) and B = x 2 , y 2 ) be two points near and on the same side of a charged sheet with surface charge density + σ . The electric field E due to such a charged sheet has magnitude E = σ 2 0 everywhere, and the field points away from the sheet, as shown in the diagram. (a) What is the potential difference V AB = V A V B between points A and B ? (b) If the potential at y = ±∞ 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? (c) Now take the potential to be zero at y = 0 instead of at infinity. What is the value of V A at point A some positive distance y 1 from the sheet? (a) Points A and B are on the same side of the sheet, so V A V B = A B E · d l = E ( y 2 y 1 ) The potential difference is positive ( V A V B > 0) because the line integral is evaluated in the direction opposite to the electric field lines. One can also say that it takes positive work to move a positive charge from B to A because one must push against the field. (b) If we move point B to y 2 = , where V B = 0, then the formula above gives V A = . (c) If we move point B to y 2 = 0, where V B = 0, then the formula above gives V A = y 1 .
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