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# ch24_p103 - 103(a The net potential is V = V1 V2 = q1 q2 4o...

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1 03. (a) The net potential is V = V 1 + V 2 = q 1 4 πε o r 1 + q 2 4 πε o r 2 where r 1 = x 2 + y 2 and r 2 = ( x d ) 2 + y 2 . The distance d is 8.6 nm. To find the locus of points resulting in V = 0, we set V 1 equal to the (absolute value of) V 2 and square both sides. After simplifying and rearranging we arrive at an equation for a circle: y 2 + © ¨ § ¹ ¸ · x + 9 d 16 2 = 225 256 d 2 . From this form, we recognize that the center of the circle is –9 d /16 = – 4.8 nm. (b) Also from this form, we identify the radius as the square root of the right-hand side: R = 15 d /16 = 8.1 nm. (c) If one uses a graphing program with “implicitplot” features, it is certainly possible to
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