PHYSICS 100B SELF-TEST PROBLEMS (1)

1.Two particles with +3.0mC and – 2.0mC charges respectively are located as shown in Fig 1. Find (a) the electric field at point P due the two charges,  (b) the electric force exerted on the 1.5mC-charge, which is placed at P.

Solution. (a) We choose the coordinate system as shown.                We find the electric field due to charge +3.0mF from                         E1 =kQ1/r12 = (9.0 x 109N.m2/C)(3.0 x 10-6C)/[(4.0m)2 + (4.0m)2];                              = 8.44 x 10N2/C.               We find the electric field due to charge -2.0mC from                         E2 = kQ2/r22 = (9.0 x 109N.m2/C)(2.0 x 10-6C)/[(2.3m)2 + (4.0m)2]                             = 8.46 x 102N/C.               We resolve E1 and E2 into x, and y-components:                        E1x = E1cos45° = (8.44 x 102N/C)cos45° = 5.97 x 102N/C;                        E1y = E1sin45° = (8.44 x 102N/C)sin45° = 5.97 x 102N/C;                        E2x = E2cos60° = (8.46 x 102N/C)cos60° = 4.23 x102N/C;                        E2y = -E2sin60° = -(8.46 x 102N/C)sin60° = -7.29 x 102N/C.                Thus we find the x-component of the resultant field from                         Ex = E1x + E2x = 5.97 x 102N/C + 4.23 x 102N/C = 1.02 x 103N/C;                         Ey = E1y + E2y = 5.97 x 10N/C - 7.29 x 10N/C = - 1.32 x 102N/C.                We find the magnitude of E from                         E = [(1.02 x 103N/C)2 + (-1.32 x 102N/C)2]1/2 = 1.03 x 103N/C                We find the angle that E with the x-axis from                        tan q = Ey/Ex = (1.33 x 102N/C)/1.02 x 103N/C), which gives                        q = 7.43° (below + x-axis).               (b) We find the force from                        F = Eq = (1.03 x 103N/C)(1.5 x 10-6C) = 1.55 x 10-3N. (along the direction of the electric field)

 

2. A 1.5mF-parallel capacitor is charged to 9mC. The air gap between the two plates of the capacitor is 0.2mm. What is the electric field in the gap? Assume that the electric field in the gap is uniform.

          Solution.  We find the potential difference between the two plates of the capacitor from

                                   Q = CV

                                   9mC = (1.5mF)V, which gives V = 6V.

                           Because the electric field between the plates is uniform we find the electric field from

                                  E = V/d = (6V)/(0.2 x 10^-3m) = 3 x 10V/m (N/C)

3. A uniform copper wire is 150m long with a circular cross sectional area, whose diameter is 0.5mm. (a) If this wire is connected to a 6V battery what is the current flows in the wire? (b) what is the power dissipated in the wire?  Taking the resistivity of copper as 1.68 x 10^-8W.m.

           Solution. (a) We find the resistance of the wire from

                                    R = rL/A = (1.68 x 10^-8W.m)(150m)/(p)(0.25 x 10m^-3)² = 12.83W.

                               Applying Ohm's Law we find the current from

                                    I = V/R = (6V)/(12.83W) = 0.468A.

                            (b) We find the power from

                                   P = I²R = (0.468A)²(12.83W) = 2.81W.

4. Calculate the current in the circuit as shown in Fig. 2.       

Solution. We combine 6W and 3W resistors into one from                          Rp = (6W)(3W)/(6W + 3W) = 2W.               We combine Rp and 1W into one from                          Rresultant = 1W + 2W = 3W.               Applying Ohm's Law we find the current from                         I = V/Rresultant = (6V)/(3W) = 2A.