# ch24_p104 - This agrees with the Rutherford field...

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1 04. The electric field (along the radial axis) is the (negative of the) derivative of the voltage with respect to r . There are no other components of E in this case, so (noting that the derivative of a constant is zero) we conclude that the magnitude of the field is E = dV dr = Ze 4 πε o © ¨ § ¹ ¸ · d r 1 dr + 0 + 1 2 R 3 d r 2 dr = Ze 4 πε o © ¨ § ¹ ¸ · 1 r 2 r R 3 for r R
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Unformatted text preview: . This agrees with the Rutherford field expression shown in exercise 37 (in the textbook). We note that he has designed his voltage expression to be zero at r = R . Since the zero point for the voltage of this system (in an otherwise empty space) is arbitrary, then choosing V = 0 at r = R is certainly permissible....
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