Download 2 nC

Physics 192 Solutions to Mastering Physics Week 8
P20.11. Prepare: We need to solve Coulomb’s law (Equation 20.1) for r :
F= K
| q1|| q2 |
r2
where F = 8 .2 × 10 −4 N and q1 = − 5 . 0 nC, q2 = − 12 nC, and K = 9 . 0 × 109 N ⋅ m 2 /C 2.
Solve:
| q || q |
r2 = K 1 2
F
r= K
| q1|| q2 |
| − 0.5 × 10 −9 C|| − 12 × 10−9 C|
9
= (9 .0 × 10 N ⋅ m 2 /C2 )
= 0 . 026 m = 2 .6 cm
F
8.2 × 10 −4 N
Assess: Notice the N and C cancel out leaving units of m. Comparing with Problem 20.10, the answer of 2.6 cm seems
to be in the right ballpark.
P20.15. Prepare: Please refer to Figure P20.15. The charged particles are point charges. The charge q2 is in static
equilibrium, so the net force on q2 is zero. If q2 is positive, q1 will have to be positive to make the net force zero on q2.
And, if q2 is negative, q1 will still have to be positive for q2 to be in equilibrium. We will assume that the charge q2 is
positive. For this situation, the force on q2 by the –2 nC charge is to the left and by q1 is to the right.
Solve: We have
 1
  1 (2 × 10−9 C)| q2|

| q1|| q2|
Fnet on q2 = Fq1 on q2 + F−2 nC on q2 = 
, + x-direction  + 
, − x-direction  = 0 N/C
2
2
 4πε 0 (0.2 m)
  4πε 0 (0.10 m)

Thus,
q1
2 × 10−9 C
−
= 0 N/C ⇒ q1 = 8.0 nC
2
(0.2 m)
(0.10 m) 2
Assess: If the charge q2 is assumed negative, the force on q2 by the –2 nC charge is to the right and by
q1 is to the left. The magnitude of q1 remains unchanged.
P20.25. Prepare: The electric field is that of the two charges placed on the y-axis. Please refer to Figure P20.25. We
denote the upper charge by q1 and the lower charge by q2. The electric field at the dot due to the positive charge is directed
away from the charge and making an angle of 45° below the +x axis, but the electric field due to the negative charge is
directed toward it making an angle of 45° below the –x axis.
Solve: The electric field strength of q1 is
E1 = K
| q1| (9.0 × 109 N ⋅ m2/C2 )(1 × 10 −9 C)
=
= 1800 N/C
(0.050 m) 2 + (0.050 m) 2
r 12
Similarly, the electric field strength of q2 is
E2 = K
| q2| (9.0 × 109 N ⋅ m 2/C2 )(1 × 10−9 C)
=
= 1800 N/C
(0.050 m) 2 + (0.050 m) 2
r 22
We will now calculate the components of these electric fields. The electric field due to q1 is away from q1 in the fourth
quadrant and that due to q2 is toward q2 in the third quadrant. Their components are
E1 x = E1 cos 45 °
E1 y = −E1 sin 45 °
E2 x = −E2 cos 45 °
E2 y = −E2 sin 45 °
The x and y components of the net electric field are:
( Enet ) x = E1x + E2 x = E1 cos 45 ° − E2 cos 45 ° = 0 N/C
( Enet ) y = E1 y + E2 y = − E1 sin 45 ° − E2 sin 45 ° = − 2500 N/C
⇒ Enet at dot = (2500 N/C, along − y axis)
Thus, the strength of the electric field is 2500 N/C and its direction is vertically downward.
Assess: A quick visualization of the components of the two electric fields shows that the horizontal components
cancel.
P20.47. Prepare: The charges are point charges. Please refer to Figure P20.47. The electric force on charge q1 is the
vector sum of the forces F2 on 1 and F3 on 1, where q1 is the 1 nC charge, q2 is the left 2 nC charge, and q3 is the right 2 nC
charge.
Solve: We have
 K | q1|| q2|

F2 on 1 = 
, away from q2 
2
r


 (9.0 × 109 N ⋅ m2/C2 )(1 × 10 −9 C)(2 × 10 −9 C)

=
, away from q2 
−2
2
(1 × 10 m)


−4
= (1.8 × 10 N, away from q2 )
( F2 on 1 ) x = (1.8 × 10−4 N)(cos 60 °) = (0.9 × 10 −4 N)
( F2 on 1 ) y = (1.8 × 10−4 N)(sin 6 0 °) = (1.56 × 10−4 N)
 K | q1|| q3|

F3 on 1 = 
, away from q3  = (1.8 × 10 −4 N, away from q3 )
r2


( F3 on 1 ) x = −(1.8 × 10 −4 N)(cos 60 °) = − (0.9 × 10 −4 N)
( F3 on 1 ) y = (1.8 × 10−4 N)(sin 60 °) = (1.56 × 10−4 N)
( Fon 1 ) x = ( F2 on 1 ) x + ( F3 on 1 ) x = 0
( Fon 1 ) y = ( F2 on 1 ) y + ( F3 on 1 ) y = 3.12 × 10 −4 N
So the force on the 1 nC charge is 3.1 × 10–4 N directed upward.
Assess: The magnitude and symmetry of q2 and q3 ensure that their x-component of the net force is zero.
P20.62. Prepare: The suspended bead is not accelerating so the net force on it must be zero. Use a free-body diagram
and Newton’s second law in both directions. The electric force is purely horizontal.
Solve:
Σ Fy = T cos 45 ° − mg = 0 ⇒ T =
Σ Fx = k
q=
mg
cos 45 °
q2
r2
2
−
T
sin
45
°
=
0
⇒
q
=
T sin 45 °
r2
k
r2
T sin 45 ° =
k
(0 .050 m)
r 2  mg 
mg
sin 45 ° = r
tan 45 ° =
k  cos 45 ° 
k
(0 . 020 × 10−3 kg)(9 .8 m/s 2 )
(1) = 7 . 4 nC
9 .0 × 109 N⋅ m 2 / C 2
Assess: This is a typical charge for a plastic bead.