Lesson_2.4

# Lesson_2.4 - Lesson 2.4 Electric Field due to Continuous...

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Lesson 2.4 Electric Field due to Continuous Charge Distribution using Gauss’s law 1. Field at a point due to an infinite plane charge. + + + + + + + + + + + + + + + + + + + + + + A E E Fig. 1 E is constant and perpendicular to the surface of the pillbox Fig. 1 shows an infinite plane charge of charge per unit area (surface charge density) σ . Field lines leave the upper as well as the lower surface of the plane surface and are perpendicular to the surface. The field has the same magnitude but opposite direction at points the same distance above and below the plane. In order to find the field at a point, we choose a Gaussian surface which encloses a small portion of the charge carried by the plane . The Gaussian surface we have chosen is in the form of a pill box shaped cylinder. Since the curved surface of the cylinder is parallel to the field lines, there is no flux crossing the curved surface. If A is the area of the face of the cylinder parallel to the plane, and E, the electric field perpendicular to it, the flux crossing one of the surfaces is E A. Since flux leaves at both faces of the cylinder, the net flux leaving the cylindrical surface is given by φ net = 2 A E Since A is the area of the plane charge inside the Gaussian surface, the net charge inside the surface = A σ 0 0 0 0 1 2 1 1 2 net inside Q A AE A E φ σ ε = = = =

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Thus field near an infinite plane charge is 0 2 2 E k σ π σ ε = = …1 Example 1 : An infinite plane of surface charge density σ = +4.5 nC.m -2 lies in the yz plane, and a second infinite plane of surface charge density σ = -4.5 nC.m -2 lies in a plane parallel to the yz plane at x = 2.0 m. Find the electric field at (a) x = 1.8 m and (b) at x = 5.0 m. Solution: + + + + + + + + - - - - - - - - - x y z Fig. 2 You notice from the figure above that the field lines start at the positive plane and end at the negative plane. Therefore, in the space between the two planes, the field due to the positive plane is in the same direction as the field due to the negative plane. Therefore, between the two planes these two fields add up. The field near an infinite plane charge is given by: 0 2 E = . Therefore, the field at any point between the two plates is: 0 0 2 2 E = + 0 = . To the left of the positive plate, the field due to the positive plate is opposite to the field due to the negative plate and so they cancel each other. Similarly, to the right of the negative plate, the field due to the positive plate cancel the field due to the negative plate. (a) At any point 0 < x < 2, the field is the same and is equal to 9 1 12 0 4.5 10 8.85 10 508.5 . N C E - - - × = = = × (b) At x = 5.0 m, E xnet = E 1 – E 2 = 0
2. Field due to a spherical shell of charge. R

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## This note was uploaded on 05/01/2011 for the course PHY 2049 taught by Professor George during the Spring '11 term at Edison State College.

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Lesson_2.4 - Lesson 2.4 Electric Field due to Continuous...

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