Physics 112 Prof' HydeWright

1
Physics 112Prof. Hyde-Wright

• Homework 1
• Solution

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1Walker1 19.P.005. (121147)

• A system of 1490 particles, all of which are
  electrons or protons,
• has a net charge of -2.528 10-17 C.
• How many electrons are in this system? Ne Np
  1490
• (Np Ne)(1.6E-19 C) -2.528 E-17 C
• 824 electrons
• (b) What is the mass of this system? MNp Mp
  Ne Me
• 1.11e-24 kg

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2 Walker1 19.P.012. (121150)
q2-2q
q1q
Q33q
Given that q -13 µC and d 13 cm, find the
direction and magnitude of the net electrostatic
force exerted on the point charge q1 in Figure
19-30. F12(force on q1 from q2) k
(q1)(q2)/(dd) (8.99E9 N mm/CC)
(-13E-6C)(26E-6C)/(0.13m)2 -180 N Minus sign
means towards q2 (attractive) F13 (force on q1
from q3) k(q2)(q3)/(2d)2 (8.99E9 N mm/CC)
(-13E-6C)(-39E-6C)/(0.26m)2 67.4 N Plus sign
means away from q3 (repulsive) Net force (180
67.4)112 N Towards q2
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3Walker1 19.P.016. (121151)
A point charge q -0.69 nC is fixed at the
origin. Where must a proton be placed in order
for the electric force acting on the proton to be
exactly opposite to its weight? F k (-0.69E-9
C) (1.6E-19C)/r2 mg Solve for r 7780 m  
Gravity pulls proton down. Proton is attracted
to q (negative charge) Place proton below q
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4Walker1 19.P.018. (121152)
Find the direction and magnitude of the net
electrostatic force exerted on the point charge
q2 in Figure 19-31. Let q 1.4 µC and d 16
cm. (Let the x axis point to the right.)
174.6 (counterclockwise from the x axis is
positive) 6.1 N Discussion in class.
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5Walker1 19.P.030. (121155)

• Two point charges lie on the x axis. A charge of
  -9.5 µC is at the origin, and a charge of 2.5
  µC is at x 10.0 cm.
• What is the net electric field at x -2.0
  cm?Electric field from charge 9.5E-6C is
• E1 (8.99E9 Nm2/C2)(-9.5E-6C)/(0-(-0.02m))2
• E1 -2.14E8 N/C
• Minus sign means field points towards charge
  9.5E-6 (positive x-direction)
• Electric field at x-0.02 m from charge 2.5E-6 C
  is
• E2 (8.99E9 Nm2/C2)(2.5E-6C)/(0.10m-(-0.02m))2
• E2 1.56E6 N/C
• Plus sign means field points away from charge
  2.5E-6 (negative x-direction)
• Net electric field E 2.14E8 i N/C - 1.56E6 i
  N/C
• 2.12e08 i N/C

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5-b
(b) What is the net electric field at x 2.0
cm? Electric field from charge 9.5E-6C is E1
(8.99E9 Nm2/C2)(-9.5E-6C)/(0-(0.02m))2 E1
-2.14E8 N/C Minus sign means field points
towards charge 9.5E-6 (negative
x-direction) Electric field at x 0.02 m from
charge 2.5E-6 C is E2 (8.99E9
Nm2/C2)(2.5E-6C)/(0.10m-0.02m)2 E2 3.51E6
N/C Plus sign means field points away from charge
2.5E-6 (negative x-direction) Net electric field
E -2.14E8 i N/C - 1.56E6 i N/C E-total-2.17e
08 i N/C
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6 Walker1 19.P.031. (121156)

• An object with a charge of -3.2 µC and a mass of
  0.036 kg experiences an upward electric force,
  due to a uniform electric field, equal in
  magnitude to its weight.
• Find the direction and magnitude of the electric
  field. Electric force qE (-3.2E-6C) E
• Since Force is upward, Electric field must be
  downward since charge lt0qEmg, Emg/q
  1.10e05 N/C
• (b) If the electric charge on the object is
  quadrupled while its mass remains the same, find
  the direction and magnitude of its acceleration.
  upward a F(net)/m F(electric)F(gravity)
  
• a (4q)E mg/m 4(qE)-mg/m 4 mg-mg/m
  3g 29.4 m/s2

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7, Walker1 19.P.050. (196953)

• A proton is released from rest in a uniform
  electric field of magnitude 1.26 105 N/C. Find
  the speed of the proton after it has traveled
  each of the following distances.
• 8.00 cm DK W -DU - (Uf-Ui) qE(xf-xi)
• Mv2/2 0 (1.6E-19 C)(1.26E5 N/C)
  (0.08m)1.61E-15 J
• V1.39e06 m/s
• (b) 80.0 cm 4.39e06 m/s

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8. Walker1 19.P.059. (121164)
Two small plastic balls hang from threads of
negligible mass. Each ball has a mass of 0.13 g
and a charge of magnitude q. The balls are
attracted to each other, and the threads attached
to the balls make an angle of 20.0 with the
vertical, as shown in Figure 19-36. Figure
19-36 (a) Find the magnitude of the electric
force acting on each ball. 0.000464 N(b) Find
the tension in each of the threads. 0.00136
N(c) Find the magnitude of the charge on the
balls. 4.66e-09 C