Solutions to Week 2 Assignment

1. For the uniform electric field between two large, parallel plates, we have

                  E = DV/d;

                  640V/m =DV /(11.0 x 10^-3m), which gives  DV = 7.04V.

2. The data given are the kinetic energies, so we find the speed from

    (a) KE_a = 1/2mv_a²;

         (750eV)(1.6 x 10^-19J/eV) = 1/2(9.11 x 10^-31kg)v_a², which gives v_a = 1.62 x 10⁷m/s.

   (b)  KE_b = 1/2mv_b²;

        (3.5 x 10³eV)(1.6 x 10^-19J/eV) = 1/2(9.11 x 10^-31kg)v_b², which gives v_b = 3.5 x 10⁷m/s.

3. (a) We find the electric potential of the proton from

                         V = kq/r = (9.0 x 10⁹N·m²/C²)(1.6 x 10^-19C)/(2.5 x 10^-15m)

                            = 5.8 x 10⁵V.

    (b) we find the electric potential energy of the system by considering one of the charges to be at the potential created by the other charge:

                       PE = qV = (1.60 x 10^-19C)(5.76 x 10⁸V) = 9.2 x 10^-14J = 0.58MeV.

4. When the electron is far away, the potential from the fixed charge is zero.

    Because energy is conserved, we have

                       DKE  + DPE =  0;

                       1/2mv² - 0 + (-e)(0 - V) = 0, or   

                      1/2mv²  = e(kQ/r)

                      1/2(9.11 x 10^-31kg)v² = (1.6 x 10^-19C)(9.0 x 10⁹N·m²/C²)(-0.125 x 10^-6C)/(0.725m),

     which gives v = 2.33 x 10⁷m/s.

5 We find the dipole moment from

                      p = el = (1.6 x 10^-19C)(0.53 x 10^-10m) = 8.5 x 10^-30C·m.

6. The uniform electric field between the plates is related to the potential difference across the plates:

                     E = V/d.

    For a parallel-plates capacitor, we have

                    Q = CV = CEd;

                    0.775 x 10^-6C = C(9.21 x 10⁴V/m)(1.95 x 10^-3m), which gives C = 4.32 x 10^-9F.

  We find the area of the plates from

                   C = Ke₀A/d;

                   4.32 x 10^-9F = (3.75)(8.85 x 10C²/N·m²)A/(1.95 x 10^-3m), which gives A = 0.254m².

7. (a) The radius of the pie plate is 

                   r = 1/2(9.0in)(2.54 x 10^-2m/in) =0.114m.

          If we assume that it approximates a parallel-plate capacitor, we have

                   C = e₀A/d = e₀pr²/d

                      = (8.85 x 10C²/N·m²)p(0.114m)²/(0.10m)

                      = 3.6 x 10^-12F = 3.6pF.

   (b) We find the charge on each plate from

                  Q = CV = (3.6pF)(9.0V) = 32pC.

  (c) We assume that the electric field is uniform, so we have

                  E =V/d = (9.0V)/(0.10m) = 90V/m.

  (d)  The work done by the battery is the energy stored in the capacitor:

                 W = U =1/2CV² = 1/2(3.6 x 10^-12F)(9.0V)² = 1.5 x 10^-10J

  (e) Because the battery is still connected, the electric field will not change. Insertion of the dielectric will change capacitance, charge, and work done by the battery.