AverageElectricField

AverageElectricField - Average Electric Field due to...

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Unformatted text preview: Average Electric Field due to Charges in a Sphere Ian R. Gatland, Georgia Institute of Technology Phys 3122 Notes, June 22, 2010 Consider a charge q on the z ­axis at a distance z above the origin inside a sphere of radius R. The average electric field in the sphere due to this charge is ˆ 3 2π π R q ( r − zz ) 2 E av = r dr sin θdθdϕ 3∫∫∫ ˆ 4 πR 0 0 0 4 πε 0 r − zz (1) where ˆ ˆ ˆ r = r sin θ cos ϕx + r sin θ sin ϕy + r cosθz (2) ˆ r − zz = ( r 2 − 2 rz cos θ + z 2 )1 / 2 (3) € From equation (2) € Combining equations (1), (2), and (3) we have E av = 2π 3q €∫ 16π 2ε 0 R 3 0 πR ∫∫ 00 ˆ ˆ ˆ [ r sin θ cosϕx + r sin θ sin ϕy + ( r cosθ − z) z ] 2 r dr sin θdθdϕ 2 2 3/2 ( r − 2 rz cos θ + z ) (4) The ϕ integration yields zero for the x and y terms in the square bracket and 2π for the z term so € E av = ˆ 3qz 8πε 0 R 3 πR ∫ ∫ (r 2 00 ( r cosθ − z) 2 2 3 / 2 r dr sin θdθ − 2 rz cos θ + z ) (5) The θ integration provides E av = € ˆ 3qz 8πε 0 R 3 Ⱥ Ⱥπ 2 ( z cosθ − r) ∫ Ⱥ z 2 (r 2 − 2rz cosθ + z 2 )1/ 2 Ⱥ r dr Ⱥ 0 0 Ⱥ R (6) which evaluates to E av = € ˆ 3qz 8πε 0 R 3 R Ⱥ ( −z − r) ∫ Ⱥ Ⱥ 0 r+z − ( z − r) Ⱥ 2 Ⱥ r dr r − z Ⱥ (7) The first term in the square brackets is –1. The second term is 1 when r < z and –1 when r > z. Thus the difference is –2 when r < z and 0 when r > z. So the integral reduces to € ˆ ˆ 3qz z qzz 2 E av = 3 ∫ ( −2) r dr = − 8πε 0 R 0 4 πε 0 R 3 € (8) ˆ The factor qzz is just the dipole moment for the single charge considered and the other factors are independent of this charge and its position. Thus, if there are several charges (or a charge distribution), the average electric field due to all the charges in the sphere is € p E av = − (9) 4 πε 0 R 3 where p is the total dipole moment of the charges. € € ...
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