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X1sol061102

# X1sol061102 - Solutions to PHY2054 Exam 1 11 June 2002 1...

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Solutions to PHY2054 Exam 1, 11 June 2002 1. Draw the force vectors at x. F(-q) is attractive to left & upward; F(+q), repulsive to right & upward. The resultant is up = +y. 2. Draw the force vectors on q as in #1, above. The resultant is to the right and parallel to a. 3. Sketch in the E-field lines for the (Q, -Q) system. A line drawn from q to the center of a is perpendicular to the field lines, so it must be an equipotential. Thus no work is done in moving along this path so W = 0! 4. Draw the force vectors or you won't follow the argument! If q = 2Q, the force due to +Q must be F(+Q) = k [Q(2Q)]/(a cos 60) 2 . But cos 60 = ½, so (a cos 60) 2 = a 2 /2 ; thus F(+Q) = 4kQ 2 /a 2 , directed upward at 60 ° to the horizontal. The force due to (-Q) has same magnitude, directed downward at 60 ° . Adding these vectors, The resultant force thus is 2 [4kQ 2 /a 2 ] cos 60 = 4kQ 2 /a 2 5. The internal r is just the (on-off potential difference) / (current) = (5.6V - 4V)/ (0.4 A) = 4 [See Text, Eqn. 18.1, 18.2, etc] 6. 'Hot' R = ('on' voltage)/ I = 4V/ 0.4A = 10 7. P = ( V) I = (4V) 0.4 A = 1.6 W 8. The KE in eV of a charge |e| accelerated through potential

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