QUIZ 1

1. A charge q1 = 7.0mC is located at the origin, and a second charge q2 = -5.0mC is located on the x-axis 0.30m from the origin (Fig.). Find the electric field at the point P, which has coordinates ( 0, 0.40) m.

Solution. 

 First , let us find the magnitude of the electric field due to each charge. The field E1 due to the 7.0-mC charge and E2 due to -5.0-mC charge are shown in Fig. Their magnitudes are

           E1 = k|q1|/r12 = (9.0 x 109N.m2/C2)(7.0 x 10-6C)/(0.40m)2 = 3.9 x 105N/C.

          E2 = k|q2|/r22 = (9.0 x 109N.m2/C)(5.0 x 10-6C)/(0.50m)2 =  1.8 x 105N/C

  The vector E1 has only a y component. The vector E2 has an x component given by E2cosq = 3/5E2 and a negative y component given by -E2sinq = -4/5E2. Hence we can express the vector as

          E1 = 3.9 x 105jN/C

          E2 = (1.1 x 105i  - 1.4 x 105j)N/C

 The resultant field E at P is the superposition of E1 and E2:

         E = E1 + E2 = (1.1 x 105i + 2.5 x 105j)N/C.

         E = [(1.1 x 105N/C) + (2.5 x105N/C)] = 2.7 x 105N/C

        q = tan (2.5 x 10N5/C)/(1.1 x 105N/C) = 66°

  
 2. A uniformly charge insulating rod of length 14cm is bent into the shape of a semicircle as in Fig. If the rod has a total charge of -7.5mC, find the magnitude and direction of the electric field at O, the center of the semicircle.

Solution

 Let l be the charge per unit length,

 Then     dq = ds = lrdq      and          dE = kdq/r2

 In component form,

              Ey = 0,   (from symmetry)

             dEx = dEcosq

  Integrating           Ex =  dEx =  ò klrcosqdq/r2

  and                      Ex = kl/r ò -p/2p/2cosqdq = 2kl/r

  But Qtotal = ll, where l = 0.14m, and r = l/p.

   Thus,                 Ex = 2pkQ/l2 = (2p)(9.0 x 109N.m2/C2)(-7.5 x 10-6C)/(0.14m)2

                            E = (-2.2 x 107N/C)i