Simple Nature - Light and Matter

shown in the figure. The length, L, of the surface is irrelevant.The field is parallel to the surface on the end caps, and thereforeperpendicular to the end caps’ area vectors, so there is no contributionto the flux. On the long, thin strips that make up the rest of thesurface, the field is perpendicular to the surface, and therefore parallelto the area vector of each strip, so that the dot product occurringin the definition of the flux is E j · A j = |E j ||A j || cos 0 ◦ = |E j ||A j |.Gauss’ law gives4πkq in = ∑ E j · A j4πkλL = ∑ |E j ||A j | .The magnitude of the field is the same on every strip, so we cantake it outside the sum.4πkλL = |E| ∑ |A j |In the limit where the strips are infinitely narrow, the surface becomesa cylinder, with (area)=(circumference)(length)=2πRL.4πkλL = |E| × 2πRL|E| = 2kλRField near a surface chargeAs claimed earlier, the result E = 2πkσ for the field near acharged surface is a special case of Gauss’ law. We choose a Gaussiansurface of the shape shown in figure j, known as a Gaussian pillbox.The exact shape of the flat end caps is unimportant.The symmetry of the charge distribution tells us that the fieldpoints directly away from the surface, and is equally strong on bothsides of the surface. This means that the end caps contribute equallyto the flux, and the curved sides have zero flux through them. If thearea of each end cap is A, then4πkq in = E 1 · A 1 + E 2 · A 2 ,where the subscripts 1 and 2 refer to the two end caps. We haveA 2 = −A 1 , so4πkq in = E 1 · A 1 − E 2 · A 14πkq in = (E 1 − E 2 ) · A 1 ,j / Applying Gauss’ law to aninfinite charged surface.and by symmetry the magnitudes of the two fields are equal, so2|E|A = 4πkσA|E| = 2πkσSection 10.6 Fields by Gauss’ Law 627