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Unformatted text preview: Physics 212 Classical and Modern Physics Spring 2008 Solutions for HandIn Set #2 Wolfson Chapter 2036 The force on the charge + e proton is. ~ F = e ~ E . Putting this into Newton’s 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 2056 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 = λ 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 xcomponents: 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 2068 (a) Since the charge is uniform, the line charge density is just λ = Q/L . (b) By symmetry there is no xcomponent. The y component must be positive, so the direction is upward. (c) Looking at Example 207, 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 xaxis, rather than over the whole xaxis. 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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 Spring '08
 Ladd
 Charge, Electrostatics, Force, Electric charge, Wolfson Chapter

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