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hw03 - Physics 21 Spring 2011 Solution to HW-3 22-12 Figure...

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Physics 21 Spring, 2011 Solution to HW-3 22-12 Figure 22-34 shows an uneven arrangement of elec- trons ( e ) and protons ( p ) on a circular arc of radius r = 1 . 95cm, with angles θ 1 = 31 . 0 , θ 2 = 47 . 0 , θ 3 = 39 . 0 , and θ 4 = 22 . 0 . What are the (a) magnitude and (b) di- rection (relative to the positive direction of the x axis) of the net electric field produced at the center of the arc? Let us redefine the given angles so that we are measuring them from the positive x axis: θ 1 = 31 . 0 , θ 2 = 78 . 0 , θ 3 = 119 . 0 , and θ 4 = 158 . 0 . Now, we can easily convert the positions of the particles from ( r,θ ) notation to ( x,y ) notation. The net electric field can be calculated as the vector sum of the individual electric fields of each particle: E = E 0 + E 1 + E 2 + E 3 + E 4 . Here, particle 0 is the electron on the x axis. Each particle’s electric field can be calculated using the vector equation E i = k q i ( r r i ) | r r i | 3 where r is the loca- tion of the point where we are calculating the electric field, and r i is the location of particle i .
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