1Bweek5 - Physics 1B Notes Week 5 Winter 2010 Electricity...

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1 Physics 1B Winter 2010 Electricity and Magnetism Notes Week 5© Walter Gekelman Let us do one more problem, which, in effect will introduce the next topic. What is the electric field outside of a uniformly charged sphere? Don’t memorize this for a test it is too complicated but read through this and understand it. Consider a sphere that is uniformly charged such that the total charge within it is Q. Let ρ be the uniform charge per unit volume. Because of symmetry the electric field must be radial and E = 1 4 πε 0 Q ˆ r r 2 . But to develop a great appreciation, even an admiration lets do it the hard way. For the sphere shown below: dV = r ' 2 dr 'sin θ ' d ' d φ ' d E = 1 4 0 ρ r ' 2 dr 'sin ' d ' d ' s 2 ˆ s We have to do the 3D integration. First of all since there is no φ dependence: d ' = 2 π . By symmetry only the radial component will survive the integration. To do the problem we will use the law of cosines twice to find α and .
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2 dE r = dE cos α = 2 π 4 πε 0 ρ r ' 2 dr 'sin θ ' d 'cos s 2 But : cos = s 2 + r 2 r ' 2 2 rs From the law of cosines: s 2 = r 2 + r ' 2 2 rr 'cos ' cos '= r 2 + r ' 2 s 2 2 rr ' Note as θ changes the distance s does but not r or r'. This is because we are in spherical coordinates and the r vectors start at the center of the circle. d (cos ') = sin ' d ' = sds rr ' E r = dE r = ρπ 2 0 r ' 2 dr ' sds cos rr ' s 2 E r = 2 ε 0 r ' = 0 r ' = R r ' 2 s s 2 rr ' s = r r ' s = r + r ' s 2 + r 2 r ' 2 ( ) 2 rs dr ' ds = 4 0 r 2 s = r r ' s = r + r ' r ' = 0 r ' = R r ' 1 + r 2 r ' 2 s 2 dr ' ds We can break this into two integrals: The first is : E first = 4 0 1 r 2 r ' dr ' ds = s = r r ' s = r + r ' r ' = 0 r ' = R 4 0 1 r 2 r + r ' ( ) r ' = 0 r ' = R r r ' ( ) r ' dr ' E first = 4 0 1 r 2 2 r ' 2 0 R dr ' = 2 3 4 0 1 r 2 R 3 Now the second intergral is:
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3 E sec ond = ρ 4 ε 0 1 r 2 r ' r 2 r ' 2 ( ) dr ' 1 s 2 ds but: 1 s 2 s = r r ' s = r + r ' r ' = 0 r ' = R ds = 1 s E sec ond = 4 0 1 r 2 r 2 r ' 2 ( ) r ' 0 R dr ' 2 r ' r 2 r ' 2 ( ) = 4 0 2 r 2 r ' 2 0 R dr ' = 4 0 2 3 r 2 R 3 Summing the first and second integrals: E = 1 4 πε 0 r 2 2 3 + 2 3 π R 3 = Q 4 0 r 2 since Q = 4 3 R 3 Therefore the charged sphere acts as if all the charge Q was contained in the center and thus behaves as a point charge. This is if you are you are at r>R. For r<R on the charge below you will count as we will see when we study Gauss’s law.
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This note was uploaded on 05/12/2010 for the course PHYSICS 1B 318007220 taught by Professor Gekelman during the Spring '10 term at UCLA.

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1Bweek5 - Physics 1B Notes Week 5 Winter 2010 Electricity...

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