Here is the answer I was given: ResponseDetails: Given thatthe two charges are q

Question

Here is the answer I was given: ResponseDetails:          Given thatthe two charges are q1 = 3.0*10-6 C and q2=-5.0*10-6 C      ----------------------------------------------------------------------------------     Let the unit positive charge placed at thepoint A then electric field due to charges rightwards asshown                      E = E1 + E2                          = (1/40) q1 / (0.5m)2+(1/40) q2 / (0.5m)2                           =--------------- V/m
What does the final answer come out to? I need to check myanswer

Explanation / Answer

          Given that the two charges are q1 = 3.0*10-6 C andq2 =-5.0*10-6 C ----------------------------------------------------------------------------------        The net electric fieldis             E = E1 + E2                = (1/40) q1 / (0.5m)2+(1/40) q2 /(0.5m)2                 = [ (9*109 N.m2 / C2) /(0.25 m2)  ]( q1 + q2)                =[ (9*109 N.m2 / C2) /(0.25 m2)  ]( 8.0*10-6C )                            = 288*103 N/C                             = (1/40) q1 / (0.5m)2+(1/40) q2 /(0.5m)2                 = [ (9*109 N.m2 / C2) /(0.25 m2)  ]( q1 + q2)                =[ (9*109 N.m2 / C2) /(0.25 m2)  ]( 8.0*10-6C )                            = 288*103 N/C             

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