P25_077 - V max = 40 V. (c) In view of our result in part...

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77. (a) By Eq. 25-18, the change in potential is the negative of the “area” under the curve. Thus, using the area-of-a-triangle formula, we have V 10 = Z x =2 0 ~ E · d~s = 1 2 (2)(20) which yields V =30V. (b) For any region within 0 <x< 3m , R ~ E · d~s is positive, but for any region for which x> 3mit is negative. Therefore, V = V max occurs at x =3m. V 10 = Z x =3 0 ~ E · d~s = 1 2 (3)(20)
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Unformatted text preview: V max = 40 V. (c) In view of our result in part (b), we see that now (to ±nd V = 0) we are looking for some X > 3 m such that the “area” from x = 3 m to x = X is 40 V. Using the formula for a triangle (3 < x < 4) and a rectangle (4 < x < X ), we require 1 2 (1)(20) + ( X − 4)(20) = 40 . Therefore, X = 5 . 5 m....
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This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.

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