HOMEWORK-CH 23

# HOMEWORK-CH 23 - HOMEWORK ASSIGNMENT 2 20 We set a...

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HOMEWORK ASSIGNMENT 2 20. We set a coordinate system with the origin at the center of the dipole, the x axis through the point P and the y axis through the charge + q . The fields due to the individual charges make an angle θ with the x -axis. The x -components of the individual fields sum to zero, while the y -components are equal and point in the ˆ j - direction. The net electric field is given by ( 29 ( 29 net 2 3 / 2 2 2 2 2 sin 2 2 y q qd E k k d r d r θ = = + + r since ( 29 2 2 2 sin . 2 d d r θ = + For r d , we write [( d /2) 2 + r 2 ] 3/2 r 3 so the expression above reduces to net 3 . y qd E k r r Since ˆ ( )j p qd = r , net 3 . p E k r ≈ - r r 28. We assume q > 0. Using the notation dq = λ dx . By symmetry, we conclude that all horizontal field components (due to the dq ’s) cancel and we need only sum (integrate) the vertical components. Symmetry also allows us to integrate these contributions over only half the rod (0 x L /2) and then simply double the result. In that regard we note that sin θ = y / r where 2

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