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Unformatted text preview: 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 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 = − Z 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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This note was uploaded on 11/06/2009 for the course PHYS 21 taught by Professor Hickman during the Spring '07 term at Lehigh University .
- Spring '07