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Unformatted text preview: HW 5: Physics 1B Spring 2010 C. M. Cooper, Teaching Assitant W. Gekelman, Professor April 28, 2010 Abstract This hw covers all of chapter 22. This is all Gauss’ Law, all the time. Remember, this only works for symmetric charge distributions where E is a constant (equidistant from the charge) and can be taken outside of the integral, and where E is perdendicular to the interation surface area dA to get rid of the dot product. This takes Gauss’ law down a peg or two so it becomes I E · dA = Q enc o ⇒ E = Q enc o A ˆ r (1) This equation is going to be the starting point for these homework solutions and for you for every Gauss’ law problem you encounter. o is a constant so theres just two things to nail down, Q enc and A . There’s 3 main options for A , A = 4 πr 2 for point charge/sphere 2 πrl for line of charge/cylinder A for a wall (2) There’s 2 main options for Q enc , the constant distribution Q enc = 4 3 πr 3 ρ for spheres πr 2 lρ for cylinders λl for lines Aσ for a wall (3) or, if you’re lucky, the non constant integral version and the Q enc = ZZZ ρ ( r ) dV = ZZ σ ( r ) dA = Z λ ( r ) dr (4) Remember that the Q enc is only the charge inside your Gaussian Surface of radius r and doesn’t depend on any charge outside it. Also, all this Q enc is condensed down to a point charge or line charge at the origin so you are always r away from the origin, not the charge! enjoy! 1 Question 22.9: Ion Paint? Really? The Ion Painted Sphere should be modeled as a shell of charge of radius 6.0 cm. BTWs, ion paint does exist, look up how a laser printer works. 1 1.1 (a) Find the charge inside the sphere This is a trick question! Since this is a sphere, we know A = 4 πr 2 . However, looking at the Q enc by our Gaussian sphere of r < 6 cm , we find there is none! Therefore, E = Q enc o A ˆ r = (0) o 4 πr 2 ˆ r = 0 (5) 1.2 (b) Find the charge just outside the sphere By the way, the phrase ”just outside” means that its at r = 6 cm but just outside the chargepaint, meaning we include it. Also, be careful, even though we are technically just next to the charge surface, all the enclosed charge gets condensed down to the origin so we are Gaussianly r = 6 cm away from it. Finally, ignore the sign on the charge and select the sign on the direction to give you the right answer. Therefore, E = Q enc o A ˆ r = 15 × 10 6 C 8 . 85 × 10 12 4 π (0 . 06 m ) 2 ( ˆ r ) = 3 . 8 × 10 7 ˆ rN/C (6) 1.3 (c) Find the charge 5 cm outside the sphere same as above, only now r = 6 cm + 5 cm E = Q enc o A ˆ r = 15 × 10 6 C 8 . 85 × 10 12 4 π (0 . 11 m ) 2 ( ˆ r ) = 1 . 1 × 10 7 ˆ rN/C (7) 2 Question 22.14: Electric Field of an atom 2.1 (a) What is the electric field just outside the nucleus?...
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 Spring '10
 gekelman
 Charge, Electrostatics, Electric charge, Coordinate system, qenc

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