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

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Unformatted text preview: 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 . 95 cm, with angles 1 = 31 . , 2 = 47 . , 3 = 39 . , and 4 = 22 . . 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: 1 = 31 . , 2 = 78 . , 3 = 119 . , and 4 = 158 . . Now, we can easily convert the positions of the particles from ( r, ) 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 + E 1 + E 2 + E 3 + E 4 . Here, particle 0 is the electron on the x axis. Each particles 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...
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This note was uploaded on 02/07/2011 for the course PHY 21 taught by Professor Hickman during the Spring '08 term at Lehigh University .

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

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