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1bch22

# 1bch22 - Chapter 22 Here are sample problems that...

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Chapter 22 Here are sample problems that illustrate how to find electric fields using Gauss’ law. First, we show generally how to find electric fields due to spherically symmetric and then cylindrically symmetric charge distributions. Consider first a spherically symmetric charge distribution, where all the charge is centered around a common origin. Suppose we wish to find the electric field E ( r ) at a distance r from the center of this distribution (the origin). Then we draw a Gaussian sphere of radius r centered at this origin. Let Q encl ( r ) denote the total charge enclosed within this sphere. From symmetry, the electric field E ( r ) is the same everywhere on the surface of this sphere, and is everywhere perpendicular to the surface (directed radially outward). Therefore, the electric flux through the sphere is the product of the electric field E ( r ) and the sphere’s surface area. That is, φ E = Z S E · dA = E ( r ) · 4 πr 2 = Q encl ( r ) ² 0 E ( r ) = Q encl ( r ) 4 π² 0 r 2 (1) Written as a vector, this electric field is E ( r ) = Q encl ( r ) 4 π² 0 r 2 ˆ r (2) where ˆ r is the unit vector pointing radially outward from the common origin of the distri- bution. Consider now a cylindrically symmetric charge distribution, where each unit length of charge is centered around a common central axis, and at a given radial distance, the charge remains constant along the length of this axis. Suppose we wish to find the electric field E ( r ) at a distance r from the common central axis of this distribution. Then we draw a Gaussian cylinder of radius r and length L centered about the central axis. Let encl ( r ) denote the total charge enclosed within this cylinder. From symmetry, the electric field E ( r ) is the same everywhere on the surface of this cylinder, and is everywhere perpendicular to the surface (directed radially outward from the central axis). Therefore, the electric flux through the cylinder is the product of the electric field E ( r ) and the cylinder’s surface area. That is, φ E = Z S E · dA = E ( r ) · 2 πrL = encl ( r ) ² 0 E ( r ) = λ encl ( r ) 2 π² 0 r (3) Written as a vector, this electric field is 1

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E ( r ) = λ encl ( r ) 2 π² 0 r ˆ r (4) where ˆ r is the unit vector pointing radially outward from the common central axis of the distribution. λ encl ( r ) is the charge per unit length enclosed within a radial distance r from the central axis. Now here are some sample problems illustrating how to apply these expressions to find the electric fields. The difficulty of each problem is on a scale from 1 to 10, 1 being least difficult, and 10 being most difficult. Each problem with a spherical distribution is immediately followed by an exactly analogous problem with a cylindrical distribution. In this way, I hope to demonstrate the parallels between the analyses for spherical and cylindrical distributions.
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