Lesson_2.4_Printable_PPT

Lesson_2.4_Printable_PPT - Using Gauss Law Field at a point...

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Using Gauss Law Field at a point due to an infinite plane charge. We can use Gauss’s law to find the electric field near an infinite plane charge. Consider a plane charge of surface charge density σ Field lines leave the upper as The field E has the same magnitude but opposite direction at points the same distance above and below the plane. well as the lower surface of the plane surface and are perpendicular to the surface.
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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 Since the curved surface of The Gaussian surface we have chosen is in the form of a pill box shaped cylinder. 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: Φ = E A
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Φ = E A Since flux leaves at both faces of the cylinder, the net flux leaving the cylindrical surface is given by: Φ net = 2E A Since A is the area of the plane charge inside the Gaussian surface, the net charge inside the surface = A σ Q inside = A σ
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Φ net = 2E A Q inside = A σ Gauss’s law is: inside net o Q φ ε = 2E A A ε σ = E 2 ε σ = o o The field near an infinite plane charge is: o E 2 ε σ = = 2 π k σ
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Example 2: 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. + + + + - - - - - x y The field lines start at the positive plane and end at the negative plane. σ - σ In the space between the two + + + + - - - - z planes, the field due to the positive plane is in the same direction as the field due to the negative plane. Between the two planes these two fields add up.
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Example 2: 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. + + + + - - - - - x y σ - σ The field near an infinite plane charge is: 0 2 E σ ε = The field at any point between the two plates is: + + + + - - - - z 0 0 2 2 E σ σ ε ε = + 0 σ ε = (a) The field at x = 1.8 m is: 0 E σ ε = 9 12 4.5 10 8.85 10 - - × = × = 508.5 NC -1
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(b) Find the electric field at at x = 5.0 m. + + + + + + + + - - - - - - - - - x y z 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.
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