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Unformatted text preview: Physics 212 Classical and Modern Physics Spring 2008 Solutions for Hand-In Set #2 Wolfson Chapter 20-36 The force on the charge + e proton is. ~ F = e ~ E . Putting this into Newtons 2nd law gives ~ F net = e ~ E = m~a . Since the acceleration is centripetal, we have a = a c = v 2 /R , so in terms of magnitudes eE = mv 2 /R E = mv 2 eR Now we substitute E = E b/R to get E b/R = mv 2 eR E = mv 2 be Solving for E : E = (1 . 67 10- 27 kg)(84 10 3 m/s) 2 (1 . 6 10- 19 C)(0 . 075 m) = 982 N/C Wolfson Chapter 20-56 P dq d L x r The piece of charge dq contributes d ~ E to the electric field at point P . We can find this contribution by treat- ing dq as a point charge, which gives dE = k dq r 2 in the + x direction. From the problem statement, dq = dx = ( Q/L ) dx , while from the geomtry above, x + r = L + d r = L + d- x Putting these into dE gives dE = kQ dx L ( L + d- x ) 2 Since all d ~ E point to the right, we can get the total ~ E by just summing the x-components: E = E x = Z dE = Z L kQ dx L ( L + d- x ) 2 = kQ L 1 L + d- x L = kQ L 1 d- 1 L + d = kQ L ( L + d )- d d ( L + d ) = kQ d ( L + d ) Wolfson Chapter 20-68 (a) Since the charge is uniform, the line charge density is just = Q/L . (b) By symmetry there is no x-component. The y- component must be positive, so the direction is upward. (c) Looking at Example 20-7, as the problem suggests, we find this case is identical except that our line of charge extends from- L/ 2 to L/ 2 along the x-axis, rather than over the whole x-axis. That means we can take the same integral set up, and then replace the lim- its of integration: E = E y = Z L/ 2- L/ 2 kQy dx L ( x 2 + y 2 ) 3 / 2 = kQy L x y 2 ( x 2 + y 2 ) 1 / 2...
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