Homework 3

Chapter 17

P3)How much kinetic energy will an electron gain (in joules and eV) if it falls through a potential difference of 21,000 V in a TV picture tube?

SOLUTION:
   The difference in potential (V) is 21,000 V.
   The kinetic energy gained is 21 keV = 21,000 x (1.6 x 10-19 J) = 3.36 x 10-15J.

P5)How strong is the electric field between two parallel plates 5.2 mm apart if the potential difference between them is 220V?

SOLUTION:
   For a constant electric field (E) where the displacement (d) is parallel to the field, V = E x d.
   E = V / d = 220V / 0.0052m = 4.2 x 104 V/m

P12)An alpha particle (which is a helium nucleus, Q = +2e, m = 6.64 x 10 -27kg) is emitted in a radioactive decay with KE = 5.53 Mev. What is its speed?

SOLUTION:
   KE = 5.53 Mev = 1/2 mv2
   v = (2 KE / m)1/2 = [2 ( 5.53 x 6) (1.6 x 10-19) / ( 6.64 x 10 -27)]1/2 m/s
   v = 1.63 x 10 7 m/s

P15)A +30 µC charge is placed 32 cm from an identical +30 µC charge. How much work would be required to move a +0.50 µC test charge from a point midway between them to a point 10cm closer to either of the charges?

SOLUTION:


   The work will be positive to overcome the greater effects of the 30 µC charge on the left.
   W = qV = q (Vb - Va)
   W = 0.5 µC (k x 30 µC) [(1/0.06m + 1/0.26m) - (1/0.16m + 1/0.16m)] = 1.08 J

P33)How much charge flows from a 12.0-V battery when it is connected to a 9.00-µF capacitor?

SOLUTION:
   C = Q / V
   Q = CV = (9 x 10 -6F) (12V) = 108 µC

P40)The electric field between the plates of a paper seperated (K = 3.75) capacitor is 9.21 x 10 4 V/m. The plates are 1.95 mm apart and the charge on each plate is 0.775 µC. Determine the capacitance of the capacitor and the area of each plate.

SOLUTION:


   C = Q / V = Q / ( | E | x d) = (0.775 x 10 -6) / (9.21 x 104)(0.00195) F    C = 4.32 x 10-9F
   C = eo k A / d (where eo = 8.85 x 10-12 C2/Nm2 is the permittivity of free space.
   A = (C d) / (K eo) = (4.32 x 10-9)(1.95 x 10-3) / [(3.75) (8.85 x 10-12)] m2
   A = 0.254 m2

Chapter 18

P7)A 9.0-V battery is connected to a bulb whose resistance is 1.6 ohms. How many electrons leave the battery per minute?

SOLUTION:
    V = I R (potential = current x resistance)
    Q = I t  (charge = current x time)
    Ne = Q / e = I t / e = (V / n) (t / e)
    Ne = (9.0 V)(60 s) / [(1.6 ohms) (1.6 x 10-19C)] = 2.1 x 1021 electrons per minute

P12)What is the diameter of a 1.00-m length of Tungsten wire whose resistance is 0.22 ohms?

SOLUTION:
    R = p L / A = p L / (pi x D2/4) = 4 p L / (pi x D2) (where p = 5.6 x 10-8 is the resistivity for Tungsten)
    D = [(4 p L) / (pi x R)]1/2 m
    D = 5.7 x 10-4 m = 0.57 mm = diameter

BU CAS PY 106
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