Solutions to Week 1 Assignment

1. Because all the charges and their separations are equal, we find the magnitude of the individual forces:
                    F1 = kQQ/L2 = kQ2/L2

                         = (9.0 x 109N·m2/C2)(11.0 x 10-6C)2/(0.150m)2

                         = 48.4N.

   The directions of the forces are determined from the signs of the charges and are indicated on the diagram.

   For the forces on the top of charge, we see that the horizontal components will cancel. For the net force, we have

                    F = F1cos30° + F1cos30° = 2F1cos30°

                       = 2(48.4N)cos30°

                       = 83.8N up, or away from the center of the triangle.

  From the symmetry each of the other forces will have the same magnitude and a direction away from the center of the triangle.

  The net force on each charge is 83.8N away from the center of the triangle.

Note that the sum for the three charges is zero.

  
2. The directions of the fields are determined from the signs of the charges and are indicated on the diagram. The net electric field will be to the left. We find its magnitude from

                  E = kQ1/L2 + kQ2/L2 = k(Q1 + Q2)/L2

                     = (9 .0 x 109N·m2/C2)( 8.0 x 10-6C + 6.0 x 10-6C)/(0.020m)2

                     = 3.2 x 108N/C.

Thus the electric field is 3.2 x 108N/C toward the negative charge.

  
3. The directions of the individual fields will be along the diagonals of the square, as shown. We find the magnitudes of the individual fields:

                   E1 = kQ1/(L/Ö2)2 = 2kQ1/L2.

                       = 2(9 .0 x 109N·m2/C2)(45.0 x 10-6C)/(0.60m)2 

                       = 2.25 x 106N/C.

                  E2 = E3  = E4 = kQ2/(L/Ö2)2 = 2kQ2/L2

                      = 2(9 .0 x 109N·m2/C2)(31.0 x 10-6C)/(0.60m)2

                      = 1.55 x 106N/C.

From the symmetry, we see that the resultant field will be along the diagonal shown as the x-axis. For the net field, we have

                 E = E1 + E3 = 2.25 x 106N/C + 1.55 x 106N/C = 3.80 x 106N/C.

Thus the field at the center is

         3.80 x 106N/C away from the positive charge.

  

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