# lecture11 - Physics for Scientists& Engineers 2 Chapter...

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Unformatted text preview: September 21, 2011 Physics for Scientists & Engineers 2, Chapter 24 1 September 21, 2011 Physics for Scientists & Engineers 2, Chapter 24 2 We grade out 6, meaning that one question is a bonus question! Monday in class: Show how to solve all exam problems Remember: We drop the lowest exam 0 5 10 15 20 25 30 35 40 45 50 Number of students <10 ≥ 10 ≥ 20 ≥ 30 ≥ 40 ≥ 50 ≥ 60 ≥ 70 ≥ 80 ≥ 90-100 >100 September 21, 2011 Physics for Scientists & Engineers 2, Chapter 24 3 Consider a capacitor constructed of two collinear conducting cylinders of length L e inner cylinder has radius r 1 and the outer has radius r 2 e inner cylinder has charge – q and the outer has charge + q September 21, 2011 Physics for Scientists & Engineers 2, Chapter 24 4 We apply Gauss’s Law to get the electric eld between the two cylinders using a Gaussian surface with radius r and length L as illustrated by the black lines Which we can rewrite to get an expression for the electric eld between the two cylinders E • d A ∫∫ = − E dA ∫∫ = − E 2 π rL ( ) = − q ε E = q ε 2 π rL September 21, 2011 Physics for Scientists & Engineers 2, Chapter 24 5 As we did for the parallel plate capacitor, we de ne the voltage difference across the two cylinders to be us, the capacitance of a cylindrical capacitor is Δ V = − E • d s i f ∫ = Edr r 1 r 2 ∫ = q ε 2 π rL dr r 1 r 2 ∫ = q ε 2 π L ln r 2 r 1 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ C = q Δ V = q q ε 2 π L ln r 2 r 1 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = 2 πε L ln r 2 / r 1 ( ) September 21, 2011 Physics for Scientists & Engineers 2, Chapter 24 6 Consider a capacitor constructed of two concentric conducting spheres with radii r 1 and r 2 e inner sphere has charge + q and the outer has charge - q September 21, 2011 Physics for Scientists & Engineers 2, Chapter 24 7 We apply Gauss’s Law to get the electric eld between the two spheres using a spherical Gaussian surface with radius r as illustrated by the red line Which we can rewrite to get an expression for the electric eld between the two spheres E • d A ∫∫ = EA = E 4 π r 2 ( ) = q ε E = q 4 πε r 2 September 21, 2011...
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## This note was uploaded on 02/03/2012 for the course MTH 235 taught by Professor Staff during the Fall '11 term at Michigan State University.

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lecture11 - Physics for Scientists& Engineers 2 Chapter...

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