Week 1

  1. A glass rod that has been charged to +12 nC touches a metal sphere. Afterward, the rod’s charge is +8 nC.
    a. What kind of charged particle was transferred between the rod and the sphere, and in which direction? That is, did it move from the rod to the sphere or from the sphere to the rod?
    b. How many charged particles were transferred?
    Solution.
    a. The rod’s charge reduced from +12 nC to 8 nC. As the sphere is a conductor the charged particles in the sphere are conduction electrons, they move from sphere to the rod.
    b. The rod’s charge reduced from +12 nC to 8 nC. As the sphere is a conductor the charged particles in the sphere are conduction electrons, they move from sphere to the rod.
  2. Three positive particles of charges 1.1 µC are located at the corners of an equilateral triangle of side 15.0 cm. Calculate the magnitude and direction of the net force on each particle.
    solution.
  3. Two charges, -Q0 and -3Q0 are a distance l apart. These two charges are free to move along the line passing through them both, but do not because there is a third charge nearby. What must be the magnitude of the third charge and its placement in order for the first two to be in equilibrium?
    Solution.

Week 2

  1. In a thunderstorm, charge is separated through a complicated mechanism that is ultimately powered by the Sun. A simplified model of the charge in a thundercloud represents the positive charge accumulated at the top and the negative charge at the bottom as a pair of point charges (figure). (a) What is the magnitude and direction of the electric field produced by the two point charges at point P, which is just above the surface of the Earth? (b) Thinking of the Earth as a conductor, what sign of charge would accumulate on the surface near point P? (This accumulated charge increases the magnitude of the electric field near point P.)

    Solution.
    (a) The electric field at point P is contributed by charge + 50 C, and charge –20 C. We find their electric fields at P:
    E1 = kQ/r12 = (8.99 x 109 N·m2/C2)(50 C)/(10 x 103)2 = 4.5 x 103 N/C (down).
    E2 = kQ/r22 = (8.99 x 109 N·m2/C2)(20 C)/(2 x 103)2 = 4.5 x 104 N/C (up).
    E = E2 – E1 = 4 x 104 N/C up.
    (b) As electrons move opposite to the direction of the electric field, and
    The Earth is a conductor, the electrons move away from point P, the ground
    near P is positively charged, or the positive charges accumulate around P.
  2. An infinitely long conducting cylinder sits near an infinite conducting sheet (Figure). The cylinder and sheet have equal and opposite charges; the cylinder is positive. (a) Sketch some electric field lines. (b) Sketch some equipotentiall surfaces.
    Solution.
  3. The two parallel plates in Figure are 2.0 cm apart and the electric field strength between them is 1.0 x 104 N/C. An electron is launched at a 45° angle from the positive plate. What is the maximum initial speed v0 the electron can have without hitting the negative plate?
     Solution.

    The electron is exerted an electric force toward the positive plate.

    Find the force;  F = -eE = ma.

    The electron follows a projectile path, with acceleration down, a = -eE/m.

    From the vertical motion:

    vy2 = vy02 + 2ad;

    To avoid the electron hitting the top plate vy must be zero.

    0 = vy02 + 2ad;

    (v0sin45º)2 = 2eEd/m = 2(1.602 x 10-19 C)(1.0 x 104 N/C)(0.02 m)/(9.1 x 10-31 kg),

    v0 = 1.18 x 107 m/s.    		  

  4. An electron beam is deflected upward through 3.0 mm while traveling in a vacuum between two deflection plates 12.0 mm apart (Figure). The potential difference between the deflecting plates is 100.0 kV and the kinetic energy of each electron as it enters the space between the plates is 2.0 x 10-15 J. What is the kinetic energy of each electron when it leaves the space between the plates?
     Solution.

    Use conservation of energy, Wfield = -ΔU, and the fact that the field is uniform to find Kf.

    ΔK = -ΔU 

    Kf = Ki - ΔU

         = Ki + Wfield

         = Ki+ eEΔy

         = Ki+ e(ΔV/d)Δy

         = 2.0 x 10-15 J + (1.602 x 10-19 C)(100.0 x 103 V)(0.0030 m)/0.0120 m)

         = 6.0 x 10-15 J