r 2 cm E 3 N C d 5 cm d 5 cm Surface area of sphere Use d not r because the

# R 2 cm e 3 n c d 5 cm d 5 cm surface area of sphere

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r = 2 cm E = 3 N / C d = 5 cm d = 5 cm § Surface area of sphere (Use d, not r, because the radius of the sphere that we’re calculating flux for is d!) § Putting this together r = 2 cm d = 5 cm E = 3 N / C Φ = E A cos A = 4 d 2 Φ = E (4 d 2 )(1) = (3 N / C)(4 )(0 . 05 m) 2 = 0 . 1 N m 2 / C = Φ = 0 cos = 1

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Prof. Satogata / Spring 2014 ODU University Physics 227N/232N 10 Gauss’s Law Problem 1 § A sphere of radius creates an electric field of strength at a distance from the center of the sphere. What is the electric charge on the sphere? r = 2 cm E = 3 N / C d = 5 cm
Prof. Satogata / Spring 2014 ODU University Physics 227N/232N 11 Gauss’s Law Problem 1: Solution § A sphere of radius creates an electric field of strength at a distance from the center of the sphere. What is the electric charge on the sphere? r = 2 cm E = 3 N / C d = 5 cm § Gauss’s law relates the enclosed electric charge to the total flux through a surface surrounding that charge § We calculated the flux in the last problem: § The only charge inside the sphere (Gaussian surface) is the electric charge on the sphere, so § The rest is just a calculation: Φ = 4 kq enclosed Φ = 0 . 1 N m 2 / C 2 q enclosed = q sphere q enclosed = q sphere = Φ 4 k = (0 . 1 N m 2 / C) 4 (9 10 9 N m 2 / C 2 ) q sphere = 8 . 1 10 - 13 C

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Prof. Satogata / Spring 2014 ODU University Physics 227N/232N 12 Gauss’s Law Example 1 § Let’s use Gauss’s Law to derive an earlier result a faster way § On pages 14-15 of the Jan 24 lecture , we derived the electric field from an infinitely long line of charge with charge per unit length § But this was painful – it involved an integral we had to look up. λ ~ E = 2 k λ y ˆ j ˆ i ˆ j
Prof. Satogata / Spring 2014 ODU University Physics 227N/232N 13 Gauss’s Law Example 1 § We will use Gauss’s law by drawing a closed surface around some of our charge § In particular, we want to draw a surface where the electric field is either perpendicular to or parallel to the surface everywhere on the surface § We also want the electric field to be the same magnitude over large pieces of the surface § Then we can use the simpler calculation for flux for separate pieces of the surface that we drew, and use Gauss’s Law Φ = EA for areas where E ? A (cos = 1) Φ = 0 for areas where E k A (cos = 0) Φ = E A cos

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Prof. Satogata / Spring 2014
• Summer '16
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