# ch22_p14 - x axis. Setting E net = 0 at x = 20 cm (see...

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(caused by q 1 and q 2 ) must point in opposite directions for x > 0. Given their locations in the figure, we conclude they are therefore oppositely charged. Further, since the net field points more strongly leftward for the small positive x (where it is very close to q 2 ) then we conclude that q 2 is the negative-valued charge. Thus, q 1 is a positive-valued charge. We write each charge as a multiple of some positive number ξ (not determined at this point). Since the problem states the absolute value of their ratio, and we have already inferred their signs, we have q 1 = 4 ξ and q 2 = −ξ . Using Eq. 22-3 for the individual fields, we find E net = E 1 + E 2 = 4 ξ 4 πε o ( L + x ) 2 ξ 4 πε o x 2 for points along the positive
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Unformatted text preview: x axis. Setting E net = 0 at x = 20 cm (see graph) immediately leads to L = 20 cm. (a) If we differentiate E net with respect to x and set equal to zero (in order to find where it is maximum), we obtain (after some simplification) that location: x = © ¨ § ¹ ¸ · 2 3 3 2 + 1 3 3 4 + 1 3 L = 34 cm. We note that the result for part (a) does not depend on the particular value of ξ . (b) Now we are asked to set ξ = 3 e , where e = 1.60 × 10 − 19 C, and evaluate E net at the value of x (converted to meters) found in part (a). The result is 2.2 × 10 − 8 N/C . 14. For it to be possible for the net field to vanish at some x > 0, the two individual fields...
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## This note was uploaded on 09/06/2009 for the course PHYSICS 14358 taught by Professor Bolton during the Spring '09 term at Kansas State University.

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