ch22_p54 - direction as that of bead 2, in that situation)....

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54. We make the assumption that bead 2 is in the lower half of the circle, partly because it would be awkward for bead 1 to “slide through” bead 2 if it were in the path of bead 1 (which is the upper half of the circle) and partly to eliminate a second solution to the problem (which would have opposite angle and charge for bead 2). We note that the net y component of the electric field evaluated at the origin is negative (points down ) for all positions of bead 1, which implies (with our assumption in the previous sentence) that bead 2 is a negative charge. (a) When bead 1 is on the + y axis, there is no x component of the net electric field, which implies bead 2 is on the – y axis, so its angle is –90°. (b) Since the downward component of the net field, when bead 1 is on the + y axis, is of largest magnitude, then bead 1 must be a positive charge (so that its field is in the same
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Unformatted text preview: direction as that of bead 2, in that situation). Comparing the values of E y at 0° and at 90° we see that the absolute values of the charges on beads 1 and 2 must be in the ratio of 5 to 4. This checks with the 180° value from the E x graph, which further confirms our belief that bead 1 is positively charged. In fact, the 180° value from the E x graph allows us to solve for its charge (using Eq. 22-3): q 1 = 4 πε o r ² E = 4 π ( 8.854 × 10 − 12 C 2 N m 2 )(0.60 m) 2 (5.0 × 10 4 N C ) = 2.0 × 10 − 6 C . (c) Similarly, the 0° value from the E y graph allows us to solve for the charge of bead 2: q 2 = 4 πε o r ² E = 4 π ( 8.854 × 10 − 12 C 2 N m 2 )(0.60 m) 2 (– 4.0 × 10 4 N C ) = –1.6 × 10 − 6 C ....
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This note was uploaded on 01/25/2011 for the course PHYSICS 17029 taught by Professor Rebello,nobels during the Fall '10 term at Kansas State University.

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