B. Conductors Homework Part I

100 points

** **

**16.1 Electric
Potential Energy and Potential Difference (Voltage)**

**1-4, 6-9, 10, 11, 13, 18, 24, 26, 27, 31**

** **

13: 1 kV = 1000
volts. Remember DV = DU/q_{o}
and KE = 1/2mv^{2}

18: DV =
kq/r_{B} - kq/r_{A}

26: U_{total}
= U_{12} + U_{23}…etc.
Since there are four charges, there will be SIX values for U. Put in the correct positive or negative
vales for each charge and then you’ll be able to determine the overall charge
at the end of your calculations.

27: V = Skq/r
where r = 10cm/cos 30^{o} = 11.5 cm = 0.115m. Again, if you use the correct positive and negative values as you
go along, you will come up with the correct charge at the end.

31: a) mass of electron = 9.11 x 10^{-31}kg. b)
velocity = distance/time. Solve
for time.

**16.2 Equipotential Surfaces and the Electric Field.**

**33, 34, 39, 40, 41, 43, 44, 46, 47, 48, 49, 51, 58, 60**

** **

41: Think of the Bohr model of an atom (like planets in orbit around the sun). Equipotential surfaces CLOSER to the nucleus are smaller in circumference. Also, V = kq/r. As r decreases (getting closer to the positive nucleus of the atom) the electric potential (voltage) increases)

43: Write the formula for an eV to see how this works.

47: DV = Ed - solve for d.

48: DV = Ed solve for d.

49: E is constant between the plates. To find E, use the formula E = V/d where d = 0.01m. Then solve for a voltage at a different distance (d = 0.0025m).

58: First figure out how to convert eV to volts – EASY! Then, convert eV to Joules and realize that KE has units of Joules – and with the KE, you can determine velocity.

60: Easy as falling off a log! Or as natural as falling off a “ln”

**Answers to Even Problems:**

2: Electrical
potential is the electrostatic potential energy per UNIT CHARGE. ie, V = U/q_{o}. b)
no difference.

4: a

6: b because the electron has negative charge.

10: +9.6 x 10^{-5}
J

18:a) 3.9 V b) The closer orbit is at the higher potential.

24: a) 6.0 x 10^{-19}J b)
3.8 V c) point C

26: -4.1J

34: b

40: Since V = kq/r and V is constant and r is constant, the equipotential surfaces are concentric spheres centered on the charge.

44: b

46: Yes. Electric field is a measure of the change in electric potential over a distance. Inside a conductor in electrostatic equilibrium, the electric field is zero, yet the elelctrical potential could be at a constant value.

48: 70 cm

58: a) 2.6 x 10^{4} m/s b)
8.9 x 10^{5} m/s
c) 9.8 x 10^{5} m/s

60: +2.8 V