Satogata Spring 2014 ODU University Physics 227N232N 14 Gausss Law Example 1

Satogata spring 2014 odu university physics 227n232n

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Prof. Satogata / Spring 2014 ODU University Physics 227N/232N 14 Gauss’s Law Example 1 § Now let’s figure out what closed surface to draw § Remember, we’ll want pieces that are perpendicular or parallel to the electric field § So we’ll need to have an idea of which direction the electric field points from this distribution of charge § The natural coordinate system to draw has the x direction along the line charge, and the y direction perpendicular to it. § In the x direction, there are always equal charges at any So the horizontal component of the field is zero, § In the y direction, all charge is located in the direction So the vertical component of the field is nonzero, ˆ i ˆ j dq dq E x = 0 ± x - y E y 6 = 0 E y ? E y 6 = 0 E x ? E x = 0

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Prof. Satogata / Spring 2014 ODU University Physics 227N/232N 15 Gauss’s Law Example 1 § The fields always point “out” from the line charge § So let’s draw a closed Gaussian surface: a cylinder § On the ends of the cylinder, so § On the side of the cylinder, and is constant: § So the total flux through this cylinder of radius y and length L is ˆ i ˆ j dq dq E y E y y L E ? A Φ = 0 E k A E Φ = E A (cylinder side area A side = (2 y)L) Φ total = 2 Φ end + Φ side = 0 + E A side = E (2 y ) L E y
Prof. Satogata / Spring 2014 ODU University Physics 227N/232N 16 Gauss’s Law Example 1 § We’re almost there! We’ve calculated : what is ? § It’s the charge per unit length times the cylinder length: § So now Gauss’s Law gives us the answer – no integrals! same answer as before!! ˆ i ˆ j dq dq E y E y y L Φ total q enclosed q enclosed = λ L ) E = 2 k λ y ˆ j Φ total = E L (2 y ) = 4 q enclosed = 4 k λ L E y
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