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Unformatted text preview: the charges qA and qU is the diagonal
of the rectangle, which is ( 4d )2 + d 2 according to the Pythagorean theorem, and the fact that the distance between the charges qA and q2 is 4d. The horizontal component of FAU is
FAU cos θ , which must be equal to FA2, so that we have
k qA qU ( 4d ) 2 + d 2 cos θ = k qA q2 qU or ( 4d ) 2 17 cos θ = q2
16 The drawing in the REASONING, reveals that cos θ = ( 4 d ) / ( 4d ) 2 + d 2 = 4 / 17 . Therefore, we find that qU 4 q2
=
17 17 16 or qU = ( ) 17 17
17 17
q2 =
3.0 ×10−6 C = 3.3 ×10−6 C
64
64 As discussed in the REASONING, the algebraic sign of the charge qU is negative . 26. REASONING AND SOLUTION
a. To find the charge on each ball we first need to determine the electric force acting on
each ball. This can be done by noting that each thread makes an angle of 18° with respect to
the vertical.
Fe = mg tan 18° = (8.0 × 10–4 kg)(9.80 m/s2) tan 18° = 2.547 × 10–3 N
We also know that
Fe = k q1 q2
r2 where r = 2(0.25 m) sin 18° = 0.1545 m. Now = kq
r2 2 Chapter 18 Problems Fe q =r k = ( 0.1545 m ) 965 2.547 × 10 –3 N
= 8.2 × 10 –8 C
9
2
2
8.99 × 10 N ⋅ m / C b. The tension is due to the combination of the weight of the ball and the electric force, the
two being perpendicular to one another. The tension is therefore, ( mg )2 + Fe2 ( )( ) 2 ( ) 2 8.0 × 10 –4 kg 9.80 m/s 2 + 2.547 × 10 –3 N = 8.2 × 10 –3 N ______________________________________________________________________________
T= 27. = SSM REASONING The charged insulator experiences an electric force due to the
presence of the charged sphere shown in the drawing in the text. The forces acting on the
insulator are the downward force of gravity (i.e., its weight, W = mg ), the electrostatic force
F = k q1 q2 / r 2 (see Coulomb's law, Equation 18.1) pulling to the right, and the tension T
in the wire pulling up and to the left at an angle θ with respect to the vertical as shown in the
drawing in the problem statement. We can analyze the forces to determine the desired
quantities θ and T. SOLUTION.
a. We can see from the diagram given with the problem statement that
Tx = F which gives and
Ty = W which gives T sin θ = k q1 q2 / r 2 T cosθ = mg Dividing the first equation by the second yields k q1 q2 / r 2
T sin θ
= tan θ =
T cos θ
mg
Solving for θ, we find that k q1 q2 mgr 2 θ = tan –1 (8.99 ×109 N ⋅ m 2 /C 2 )(0.600 ×10 –6 C)(0.900 ×10 –6 C) = tan –1 = 15.4°
(8.00 × 10 –2 kg)(9.80 m/s 2 )(0.150 m)2 966 ELECTRIC FORCES AND ELECTRIC FIELDS b. Since T cosθ = mg , the tension can be obtained as follows:
mg
(8.00 ×10−2 kg) (9.80 m/s 2 )
=
= 0.813 N
cos θ
cos 15.4°
______________________________________________________________________________
T= 28. REASONING AND SOLUTION Since the objects attract each other initially, one object
has a negative charge, and the other object has a positive charge. Assume that the negative
charge has a magnitude of q1 and that the positive charge has a magnitude of q2 .
Assume also that q2 is greater than q1 . The magnitude F of the initial attractive force
between the objects is F= k q1 q2 (1) r2 After the objects are brought into contact and returned to their initial positions, the charge
on each object is the same and has a magnitude of q2 − q1 / 2 . The magnitude F of the ( ) force between the objects is now ( ) k q2 − q1 / 2 F= 2
r 2 It is given that F is the same in Equations (1) and (2).
Equations (1) and (2) and rearrange the result to find that
2 q1 − 6 q2 q1 + q2 2 (2)
Therefore, we can equate =0 (3) The solutions to this quadratic equation are
q1 = 5.828 q2 and q1 = 0.1716 q2 (4) Since we have assumed that q2 is greater than q1 , we must choose the solution q1 = 0.1716 q2 . Substituting this result into Equation (1) and using the given values of
F = 1.20 N and r = 0.200 m, we find that
q1 = 0.957 µ C and q2 = 5.58 µ C (5) Note that we need not have assumed that q2 is greater than q1 . We could have assumed
that q2 is less than q1 . Had we done so, we would have found that Chapter 18 Problems q1 = 5.58 µ C q2 = 0.957 µ C and 967 (6) Considering Equations (5) and (6) and remembering that q1 is the negative charge, we
conclude that the two possible solutions to this problem are q1 = −0.957 µ C and q2 = +5.58 µ C or q1 = −5.58 µ C and q2 = +0.957 µ C ______________________________________________________________________________
29. SOLUTION Knowing the electric field at a spot allows us to calculate the force that acts on
a charge placed at that spot, without knowing the nature of the object producing the field.
This is possible because the electric field is defined as E = F/q0, according to Equation 18.2.
This equation can be solved directly for the force F, if the field E and charge q0 are known.
SOLUTION Using Equation 18.2, we find that the force has a magnitude of ( ) F = E q0 = ( 260 000 N/C ) 7.0 × 10 –6 C = 1.8 N
If the charge were positive, the direction of the force would be due west, the same as the
direction of the field. But the charge is negative, so the force points in the opposite
direction or due east. Thus, the force on the charge is 1.8 N due east .
_________...
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This note was uploaded on 04/30/2013 for the course PHYS 1100 and 2 taught by Professor Chastain during the Spring '13 term at LSU.
 Spring '13
 CHASTAIN
 Physics, The Lottery

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