Homework 2

Chapter 16

Q18)Consider the electric field at the three points indicated by the letters A, B, and C in Fig. 16-36 (of your text). First draw an arrow at each point indicating the direction of the net force that a positive test charge would experience if placed at that point, then list the letters in order of decreasing field strength (strongest first).

SOLUTION: Pictoral representation - on reserve in S and E library
For a positive charge, the force is in the same direction as the field. As far as strength goes, the closer the field lines, the stronger the field. Looking at how close the lines are at A, B, and C indicates that the field is strongest at A and weakest at C. In fact, the field is zero at C (imagine that you're a positive charge placed between these two charges - what net force acts on you?).

P22)What is the magnitude and direction of the electric force on an electron in a uniform electric field of strength 3500 N/C that points due east?

SOLUTION: Since the E electric field points east, the force on an electron would point in the opposite direction, west. (The F force on a proton would point in the same direction as theE field).
    | F | = e | E | = (1.6 x 10 -19 C) (3500 N/C)
    | F | = 5.6 x 10 -16 N

P26)What is the magnitude and direction of the electric field at a point (A) midway between a -8.0 µC and a +6.0 µC charge 4.0 cm apart?

SOLUTION:


   so the net E points toward Q1:
   | E | = | E1 | + | E2 | = [k / (0.02 m)2] (8.0 x 10-6 + 6.0 x 10-6)
   | E | = 3.2 x 108 N/C.

P29)Determine the direction and magnitude of the electric field at the point P shown in Fig. 16-40 (of your text). The two charges are seperated by a distance of 2a and the point P is a distance of x out on the perpendicular bisector of the line joining them. Express your answers in terms of Q, x, a, and k.

SOLUTION:


    E = E+Q + E-Q
   Due to symmetry, the x-coordinate components of E+Q and E-Q cancel. Thus,The y-components are equal, and they both point down, so the net field is down.
   E = 2 |EQ| sin(theta) = [ 2 (k Q) / (x2 + a2) ] [ (a / (x2 + a2)1/2 ]
   E = (2 k Q a) / (x2 + a2)3/2

P31)Calculate the electric field at the center of a square 60 cm on a side if one corner is occupied by a +45.0 µC charge and the other three are occupied by a -31.0 µC charges.

SOLUTION:


   The fields from Q2 and Q4 cancel E2 + E4 = 0, so E = E1 + E2 + E3 + E4 = E1 + E3
   | E | = | E1 | + | E3 | = (k x | Q1 |) / r2 + (k x | Q3 |) / r2 and r = (21/2) (30 cm)
   | E | = [ (9.0 x 10 9 Nm2/C2) / [(21/2) (30 cm)]2 ] x (45 µC + 31 µC)
   | E | = 3.8 x 10 6 N/C

P34)Draw, approximately, the electric field lines about two point charges, +Q and -3Q, which are a distance l apart.

SOLUTION: Pictoral representation - on reserve in S and E library
Because the negative charge is three times larger than the positive charge, three times as many field lines come in to the negative charge as leave the positive charge.

P38)An electron (mass m = 9.1 x 10 -31kg is accelerated in the uniform field E (E = 1.85 x 10 4 N/C) between the two parallel charged plates. The seperation of the plates is 1.20 cm. The electron is accelerated from rest near the negative plate, (Fig. 16-44 of your text). (a) With what speed does it leave the hole? (b) Show that the gravitational force can be ignored.

SOLUTION:
   (a)ma = F = e E so the acceleration a = (1.6 x 10-19) (1.85 x 10 4) / (9.1 x 10 -31) (m/s2).
   a = 3.25 x 10 15 m/s2
   v = (a)(t), x = 1/2 a t2, so t = [(2)(x) / (a)] 1/2 and v = [(2)(x)(a)]1/2
   v = [2(0.012)(3.25 x 10 15]1/2 (m/s) = 8.83 x 106 m/s
   (b)gravitational force magnitude = mg = (9.1 x 10 -31 kg) (9.8 m/s2) = 8.9 x 10 -30 N
   electric force magnitude = e E = 2.96 x 10 -15 N , so mg << e E and the effects due to gravity are extremely
   small.

P49)A point charge (m = 1.0 g) at the end of an insulating string of length 50 cm (Fig. 16-46) is observed to be in equilibrium in a known uniform horizontal electric field, E = 9200 N/C, when the pendulum has swung so it is 1.0 cm high. If the field points to the right in Fig. 16-46, determine the magnitude and sign of the point charge.

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SOLUTION:


since the charge is 1.0 cm above its position, cos(theta) = (L - 1 cm) / L = 49 / 50 and so, the angle theta = 11.5o
   There are 3 forces on the charge Q: electric, gravity and tension.
    | Fg | = | FT | cos(11.5o)
    and | FT | sin(11.5o) = | FE| = | Fg | tan(11.5o) = mg tan(11.5o)
   = (0.001kg) (9.8 m/s2) (0.20) = | FE |
   |FE | = | Q | | E | so | Q | = 1.96 x 10 -3 N / 9200 (N/C) = 2.1 x 10 -7 C
   Q is positive since FE is in the same direction as E

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BU CAS PY 106
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