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PPE10_UsingGausssLaw2

# PPE10_UsingGausssLaw2 - Electric Fields Using Gauss's Law...

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Electric Fields Electric Fields Using Gauss’s Law, Using Gauss’s Law, Continued Continued © RHJansen

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Solving E with Multiple Layered Spheres a b c Charges pile up on the surface of conductors, so there is no net charge at the center. + + + + + + + + + + + + + + + + © RHJansen A + q charge resides on a solid conducting sphere of radius a . Surrounding this is a hollow neutral sphere made of conducting material. Its inner radius is b and the outer radius is c . a) What is the electric field at the center ?
Solving E with Multiple Layered Spheres b c E = 0 There is no charge in the center Also a point has no volume Zero is enclosed + + + + + + + + + + + + + + + + © RHJansen A + q charge resides on a solid conducting sphere of radius a . Surrounding this is a hollow neutral sphere made of conducting material. Its inner radius is b and the outer radius is c . a) What is the electric field at the center ? a

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Solving E with Multiple Layered Spheres E = 0 First imagine a Gaussian surface at this location. b c + + + + + + + + + + + + + + + + Solve the electric field equation plugging in only the net charge enclosed by the surface. E = κ θ ρ 2 = κ (029 2 © RHJansen b) What is the electric field at a radius r that is 0 < r < a a
Solving E with Multiple Layered Spheres E = κ θ α 2 Look at Gaussian surface and net charge enclosed. b c + + + + + + + + + + + + + + + + The full + q charge of the inner sphere is inside this surface. E = κ ρ 2 = κ + θ ( 29 2 © RHJansen c) What is the electric field at a radius r where r = a ? a

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Solving E with Multiple Layered Spheres E = κ θ ρ 2 Look at Gaussian surface and net charge enclosed. b c + + + + + + + + + + + + + + + + The full + q charge of the inner sphere is inside this surface. E = κ 2 = κ + θ ( 29 ( 29 2 © RHJansen d) What is the electric field at radius r where a < r < b ? a
Solving E with Multiple Layered Spheres b c + + + + + + + + + + + + + + + + The electric field coming out of the inner sphere interacts with the charges in the outer conducting ring. Charges in a conductor can move. Negative charges move toward the inner positive sphere, and they pile up on the inner surface of the outer sphere. This leaves the outer surface of the outer sphere as positive. + + + + + + + + + + + + + + + + © RHJansen e) What is the electric field at radius r where b < r < c ? a

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Solving E with Multiple Layered Spheres b c + + + + + + + + + + + + + + + + The electric field inside the outer conducting ring vanishes. We are left with a field that extends from the positive inner ring to the negative inside surface of the outer ring. Another field extends outward from the outmost surface. + + + + + + + + + + + + + + + + © RHJansen e) What is the electric field at radius r where b < r < c ? a
Solving E with Multiple Layered Spheres The inner sphere has a + q charge. The second sphere has a negatively charged inner surface. This charge is equal in magnitude to the charge on the inner sphere, but has the opposite sign, q .

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PPE10_UsingGausssLaw2 - Electric Fields Using Gauss's Law...

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