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Unformatted text preview: - Summer Session 2 - C. Palmer August 8, 2013 20 Charge Density +λ (C/m)   The electric field points away from the line of charge.   The normal to the surface also points in that direction. ˆˆ n=z ˆˆ n=r ˆ E (d ) r L Physics 2B - Summer Session 2 - C. Palmer August 8, 2013 d 21 Charge Density +λ (C/m)   Note that the direction of the electric field being purely in this direction is contingent on the line of charge being infinite in both directions. ˆˆ n=z ˆˆ n=r ˆ E (d ) r L Physics 2B - Summer Session 2 - C. Palmer August 8, 2013 d 22 Charge Density +λ (C/m)   Gauss’s Law ∫ QEnclosed E ⋅ dA = ε0 ˆˆ n=z ˆˆ n=r ˆ E (d ) r L Physics 2B - Summer Session 2 - C. Palmer August 8, 2013 d 23 Charge Density +λ (C/m)   Gauss’s Law (Lefthand Side) ∫ E ⋅ dA = ∫ ˆˆ E ( d ) r ⋅ r dA ∫ ˆˆ E ( r ) r ⋅ z dA Si de + Top + ∫ ˆˆ n=z ˆˆ n=r ˆ E (d ) r ˆˆ E ( r ) r ⋅ ( − zdA ) L d Bottom Physics 2B - Summer Session 2 - C. Palmer August 8, 2013 24 Charge Density +λ (C/m)   Gauss’s Law (Lefthand Side) ∫ E ⋅ dA = ∫ Si de ˆˆ E ( d ) r ⋅ r dA = E ( d ) 2π dL ˆˆ n=z ˆˆ n=r ˆ E (d ) r L Physics 2B - Summer Session 2 - C. Palmer August 8, 2013 d 25 Charge Density +λ (C/m)   Gauss’s Law (Righthand side) QEnclosed Lλ = ε0 ε0 ˆˆ n=z ˆˆ n=r ˆ E (d ) r L Physics 2B - Summer Session 2 - C. Palmer August 8, 2013 d 26 Charge Density +λ (C/m)   Gauss’s Law Lλ E ( d ) 2π dL = ε0 ˆˆ n=z ˆˆ n=r λ kE λ ⇒ E (d ) = =2 ˆ 2πε 0 d d E (d ) r L Physics 2B - Summer Session 2 - C. Palmer August 8, 2013 d 27   Gauss’s Law can be used to determine the electric field through OPEN surfaces when there is some symmetry to be exploited.   Often the open surface in question can be incorporated into a larger Gaussian surface.   Choosing the larger Gaussian surface is the tough part   Rule of thumb: Pick the surface where the electric field is constant.   Then the electric field comes out of the integral Physics 2B - Summer Session 2 - C. Palmer August 8, 2013 28   We have a point charge (+Q) at d/2 above the center an imaginary open surface which is square with side length d.   We would like to know the electric flux from the point charge’s field. Physics 2B - Summer Session 2 - C. Palmer August 8, 2013 29 ΦE = ∫ Surface d 2 E ⋅ dA = ∫ d − keQ ˆˆ z ⋅ r dx dy ∫−d 2 r 2 2 d 2   Doable… but hideous Physics 2B - Summer Session 2 - C. Palmer August 8, 2013 30   Imagine now that the square is one side of a cube and the charge lies at the center of it.   The geometry works because of the dimensions chosen.  ...
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