COLLEGE PHYSICS II   TEST NO. 2   SPRING 2008

   Problems worth 15 points each.

     1. A charge of 18.5 nC is located at the origin of a coordinate system,
	(0,0).  Another charge, -12.4 nC, is originally located at the point
	(0, 0.050 m), but is then moved to a new point, (0.050 m, 0.100 m).
	What is the change in the electrical potential energy of this system
	of charges?

	The potential energy between two charges is

	PE = koq1q2 / r

	plus a constant that depends on the zero level chosen.  If infinite
	separation is the zero level, that constant is zero.  In any event
	the constant subtracts out whenever you find the difference in
	potential energy between two charges when one or both are moved. In
	this case, the initial potential energy is, taking the zero level at
	infinity,

	PEi = koq1q2 / r1,

	where r1 is the initial separation of the charges.  The
	final PE is

	PEf = koq1q2 / r2,

	where r2 is the final separation.  Note that electrical
	potential energy is not the same as electrical potential.  The latter
	is the electrical potential energy per unit charge.

	The charge at the origin does not move.  The initial and final
	positions of the other charge are shown in the following diagram.

        y

        |
        |       (-)  (0.05, 0.10)
        |
        |
       (-)  (0, 0.05)
        |
        |
       (+)------------------------- x


	From the diagram,

	r1 = 0.05 m

	r2 = [(0.05 - 0)2 + (0.10 - 0)2]1/2 m = 0.112 m

	Hence,

	PEf - PEi = (9 × 109 N·m2/C2(18.5 × 10-9 C)(-12.4 × 10-9 C) / (0.112 m) - (9 × 109 N·m2/C2(18.5 × 10-9 C)(-12.4 × 10-9 C) / (0.05 m)
	= 2.28 × 10-5 J


     2. A parallel-plate capacitor has an electric field of 500 V/m between
	its plates.  What is the magnitude of the surface charge density
	(charge per unit area) on the plates?  If the plates are 2.4 mm apart,
	what is the voltage difference between the plates?

	The electric field between a the plates of a parallel-plate
	capacitor is given by the equation

	E = σ / Keεo,

	where Ke, the dielectric constant, is one for vacuum and
	practically the same for air.  Solving for the charge per unit area,
	σ, you get

	σ = εoE = 8.854 × 10-12 C2/N·m2(500 N/C) = 4.43 × 10-9 C/m2.

	The voltage difference in an electric field is given by the formula

	ΔV = Ed,

	where d is the distance between two points as measured parallel to the
	field.  For a parallel-plate capacitor, d is just the separation of
	the plates.  Hence,

	ΔV = (500 V/m)(0.0024 m) = 1.2 V


     3. When a 12.0-V battery (that is, emf = 12.0 V) is being recharged by
	forcing a current through the battery opposite the direction of its
	emf.  At one point the current is 5.00 A and voltage across the
	battery measures 14.5 V.  What is the internal resistance of the
	battery at this point?

	The situation is as follows,


            12.0 V
              |         r
	a-----| |----/\/\/\/\------b
              |

	    ------> 5.00 A

	The voltage across the battery is the voltage difference between
	points a and b.  Going from b to a, the voltage increases across
	the internal resistance by an unknown amount and then increases
	again across the emf.  Since the emf is 12.0 V and the total
	voltage difference is 14.5 V, there must be a voltage difference of
	2.5 V across the resistance.  Then, from V = IR,

	r = 2.5 V / 5 A = 0.50 Ω


     4. A long, straight wire carries a current of 3.80 A.  If the wire is
	oriented in the north-south direction and the current is flowing to
	the north, what is the magnetic field (magnitude and direction) at a
	point 20.0 cm above the wire?  If an electron at some point in time
	is moving to the north 20.0 cm above the wire at a speed of 5.00E+07
	m/s, what will be the Lorentz (magnetic) force on the electron at
	that time?  Give both the magnitude and direction of the force, and
	ignore the Earth's magnetic field. Recall e = 1.60E-19 C.

	The diagram below illustrates the situation.

                     o -----> B

             0.200 m

                    (×) 3.80 A

	The right-hand rule with the current into the page (north) determines
	the magnetic field to be to the right (east) directly above the wire.
	Now, to find the magnitude of the field, you use

	B = μoI / 2πr = (4π × 10-7 T·m/A)(3.8 A) / 2π(0.2 m) = 3.80 × 10-6 T = 3.80 μT.

	For the second part, use the Lorentz force equation.

	FM = qv⊥B = (1.60 × 10-19 C)(5.00 × 107 m/s)(3.80 × 10-6 T) = 3.00 × 10-17 N.

	The RHR shows the force on a positive charge to be downward directly
	above the wire, therefore the electron experiences an upward force.