MIDTERM I

1. Two 2.0mC point charges are located on the x-axis. One is at x = 1.0m, and the other is at x = -1.0m. (a) Determine the electric field on the y axis at y = 0.50m. (b) Calculate the electric force on a -3.0mC chargee placed on the y-axis at y = 0.50m.

Solution.  (a) Because the two charge are equal and are symmetry to y-axis the magnitudes of the electric field due to two charges are equal:

                      E1 = E2 = Q/4peor2,  

               where r2 = (1.0m)2 + (0.5m)2 = 1.25m2, we get  

                      E1=E2 = (9 x 109Nm2/C2)(2.0 x 10-6C)/1.25m2 = 1.44 x 104N/C

                      E = E1 + E2,                 

                      E = 2Ecos q = (2)(1.44 x 104N/C)(0.5m/1.12m) 

                         =  1.29 x 104N/C

                      E = (1.29 x 104N/C)j

               (b) The force exerted on charge -3.0mC is

                       F = qE = (-3.0 x 10-6C)(1.29 x 104N/C)j = (- 3.87 x 10-2N)j

 

 

2. A solid sphere of radius 40.0cm has a total  positive charge of 26.0mC uniformly distributed throughout its volume. Calculate the magnitude of the electric field (a) 10.0cm, (b) 40.0cm, and (c) 60.0cm from the center of the sphere.

Solution.  Choose the concentric spherical surfaces with the center at the center of the solid sphere as gaussian surfaces as shown in figure. Assume the radius of the solid sphere is R.0  and the radius of the gaussian surface is R.

          (a)  At R = 10.0cm, the charge enclosed by the gaussian surface is 

                Q'= r4/3pR3 = (Q/4/3pR03)(4/3pR3) = Q(R/R0)3 = (26.0 x 10-6C)(10.0cm/40.0cm)3

                    = 4.06 x 10-7C

               Apply Gauss's law, we get

                 ò Edl = E4pR2 = Q'/e0. which gives 

                 E = Q'/4pe0R2 = (4.06 x 10-7C)(9 x 10Nm2/C2)/(10 x 10-2m)2 = 3.60 x 105N/C

                E = (3.60 x 105N/C)r    radiating away from the center.

          (b) At R = 40cm, the charge enclosed by the gaussian surface is

                Q' = Q, we get

                E = Q/4pe0R = (26.0 x 10-6C)(9 x 109Nm/C)/(40.0 x 10m)2 = 1.46 x 106N/C

                E = (1.46 x 106N/C)r     Radiating from the center.

           (c) At R =  60.0cm, we have

                Q' = Q,

                 E = (26 x 10-6C)(9 x 109Nm2/C2)/(60.0 x 10m)2 = 6.50 x 105N/C

                E = (6.50 x 105N/C)r    radiating from the center.

 

3. A rod of length L lies along the x-axis with its left end at the origin and has a non-uniform charge density l = ax (where a is a positive constant).

    Calculate the electric potential at A.

Solution.  Consider a infinitesimal bit of the rod at location x and of length dx. The amount of charge on it is 

               ldx. Its distance from A is d + x, so the bit of electric potential it creates at A is

                           dV = (1/4pe0)dq/r = (1/4pe0)(ax)dx/(d + x)

               We must integrate all these contributions for the whole rod, from x = 0 to x = L:

                          V =  ò 0L(1/4pe0)axdx/(d + x)

               To perform the integral, make a change of variables to 

                         u = d + x, du = dx, u (at x = 0) = d, and u (at x = L) = d +  L

                        V =  ò dd+L(4pe0)a(u - d)du/u = (1/4pe0)[a ò dd+Ldu - ad ò dd+L(1/u)du]

                           = (a/4pe0)[u|dd+L - dlnu|dd+L

                                    = (a/4pe0)[(d + L -d) - d(ln(d + L) - lnd]

                           = (a/4pe0){L - dln[(d + L)/d]}

              

  

4. (a) Determine the equivalent capacitance for the capacitor network shown in Figure 2. (b) If the network is connected to a 12-V battery, calculate the potential difference across each capacitor and the charge on each capacitor.

Solution. (a) We can get the equivalent capacitance as shown.

                    Ceq-series = (3.0 x 10-6F)(6.0 x 10-6F)/(3.0 x 10-6F + 6.0 x 10-6F) 

                                      = 2.0 x 10-6F = 2.0mF.

                   Ceq-parallel = 2.0 x 10-6F + 2.0 x 10-6F = 4.0mF.

              (b) The potential difference across 2.0mF capacitor is 12.0V,  so the charge on 2.0mF capacitor is

                       Q2.0mF = (2.0 x 10-6F)(12V) = 24.0mC.

                   The charge on the combination of 3.0mF and 6.0mF capacitors is also 24.0mC. 

                   The potential difference across 3.0mF  capacitor is

                       V3.0mF = (24.0mC)/(3.0mF) = 8.0V.

                   The potential difference across 6.0mF capacitor is

                      V6.0mF = (24.0mF)/(6.0mF) = 4.0V.