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Hwk3Solns

# Hwk3Solns - Assignment 3 Solutions The potential is VP = 1...

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Assignment 3 Solutions Chpt 24 24. . The potential is 92 2 1 2 2 rod rod 00 0 1 1 ( 8 . 9 91 0Nm C) ( 2 5 . 61 0 C ) 44 4 3 . 7 1 1 0 m 6.20 V. P dq Q Vd q RR R εε ε −× ⋅× == = = ππ π × =− ∫∫ We note that the result is exactly what one would expect for a point-charge – Q at a distance R . This “coincidence” is due, in part, to the fact that V is a scalar quantity. 30. Consider an infinitesimal segment of the rod, located between x and x + dx . It has length dx and contains charge dq = λ dx , where λ = Q / L is the linear charge density of the rod. Its distance from P 1 is d + x and the potential it creates at P 1 is x d dx x d dq dV + = + = λ πε 0 0 4 1 4 1 To find the total potential at P 1 , we integrate over the length of the rod and obtain: 0 0 0 2 1 5 3 ln( ) ln 1 4 (8.99 10 N m C )(56.1 10 C) 0.12 m ln 1 7.39 10 V. 0.12 m 0.025 m L L dx Q L x dx L d λλ ⎛⎞ + = + ⎜⎟ + ⎝⎠ ×⋅ × =+ π = × 38. (a) From the result of Problem 24-30, the electric potential at a point with coordinate x is given by 0 ln . 4 Qx L V Lx = p At x = d we obtain 2 1 5 0 3 (8.99 10 N m C )(43.6 10 C) 0.135 m ln ln 1 4 0.135 m 0.135 m (2.90 10 V)ln 1 . Qd L V Ld d d × + + π 1

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(b) We differentiate the potential with respect to x to find the x component of the electric field: 2 00 92 2 1 5 42 1 ln 44 4 (8.99 10 N m C )(43.6 10 C) (3.92 10 N m C) , ( 0.135 m) ( 0.135 m) x V Q xL Q x Q E 0 ( ) x Lx x LxLx x xxL xx εε −− ∂∂ ⎛⎞ =− ⎜⎟ ⎝⎠ ×⋅ × × ++ ππ π ε or (3.92 10 N m C) || ( 0.135 m) x E = + .
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Hwk3Solns - Assignment 3 Solutions The potential is VP = 1...

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