Physics for Scientists and Engineers: Foundations and Connections
1st Edition
ISBN: 9781133939146
Author: Katz, Debora M.
Publisher: Cengage Learning
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Chapter 39, Problem 38PQ
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The proof of
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Point charges of 6.50 μC and -2.50 μC are placed 0.300 m apart. (Assume the negative charge is located to the right of the positive charge. Include the sign of the value in your answers.)
(a) Where can a third charge be placed so that the net force on it is zero?
0.49
m to the right of the -2.50 μC charge
(b) What if both charges are positive?
0.49
xm to the right of the 2.50 μC charge
Find the electric field at the location of q, in the figure below, given that q₁ =9c9d = +4.60 nC, q = -1.00 nC, and the square is 20.0 cm on a side. (The +x axis is directed to the right.)
magnitude
direction
2500
x
What symmetries can you take advantage of? What charges are the same magnitude and the same distance away? N/C
226
×
How does charge sign affect the direction of the electric field? counterclockwise from the +x-axis
9a
9b
%
9
9d
would 0.215 be the answer for part b?
Chapter 39 Solutions
Physics for Scientists and Engineers: Foundations and Connections
Ch. 39.1 - Which of the following are (approximately)...Ch. 39.2 - Suppose the primed and laboratory observers want...Ch. 39.7 - Prob. 39.3CECh. 39.10 - Prob. 39.4CECh. 39.12 - Prob. 39.5CECh. 39 - Prob. 1PQCh. 39 - Prob. 2PQCh. 39 - Prob. 3PQCh. 39 - In an airport terminal, there are two fast-moving...Ch. 39 - Prob. 5PQ
Ch. 39 - Prob. 6PQCh. 39 - Prob. 7PQCh. 39 - Prob. 8PQCh. 39 - Prob. 9PQCh. 39 - Prob. 10PQCh. 39 - Prob. 11PQCh. 39 - Prob. 12PQCh. 39 - Prob. 13PQCh. 39 - Prob. 14PQCh. 39 - Prob. 15PQCh. 39 - Prob. 16PQCh. 39 - Prob. 17PQCh. 39 - Prob. 18PQCh. 39 - Prob. 19PQCh. 39 - Prob. 20PQCh. 39 - Prob. 21PQCh. 39 - Prob. 22PQCh. 39 - Prob. 23PQCh. 39 - A starship is 1025 ly from the Earth when measured...Ch. 39 - A starship is 1025 ly from the Earth when measured...Ch. 39 - Prob. 26PQCh. 39 - Prob. 27PQCh. 39 - Prob. 28PQCh. 39 - Prob. 29PQCh. 39 - Prob. 30PQCh. 39 - Prob. 31PQCh. 39 - Prob. 32PQCh. 39 - Prob. 33PQCh. 39 - Prob. 34PQCh. 39 - Prob. 35PQCh. 39 - Prob. 36PQCh. 39 - Prob. 37PQCh. 39 - Prob. 38PQCh. 39 - As measured in a laboratory reference frame, a...Ch. 39 - Prob. 40PQCh. 39 - Prob. 41PQCh. 39 - Prob. 42PQCh. 39 - Prob. 43PQCh. 39 - Prob. 44PQCh. 39 - Prob. 45PQCh. 39 - Prob. 46PQCh. 39 - Prob. 47PQCh. 39 - Prob. 48PQCh. 39 - Prob. 49PQCh. 39 - Prob. 50PQCh. 39 - Prob. 51PQCh. 39 - Prob. 52PQCh. 39 - Prob. 53PQCh. 39 - Prob. 54PQCh. 39 - Prob. 55PQCh. 39 - Prob. 56PQCh. 39 - Consider an electron moving with speed 0.980c. a....Ch. 39 - Prob. 58PQCh. 39 - Prob. 59PQCh. 39 - Prob. 60PQCh. 39 - Prob. 61PQCh. 39 - Prob. 62PQCh. 39 - Prob. 63PQCh. 39 - Prob. 64PQCh. 39 - Prob. 65PQCh. 39 - Prob. 66PQCh. 39 - Prob. 67PQCh. 39 - Prob. 68PQCh. 39 - Prob. 69PQCh. 39 - Prob. 70PQCh. 39 - Joe and Moe are twins. In the laboratory frame at...Ch. 39 - Prob. 72PQCh. 39 - Prob. 73PQCh. 39 - Prob. 74PQCh. 39 - Prob. 75PQCh. 39 - Prob. 76PQCh. 39 - Prob. 77PQCh. 39 - In December 2012, researchers announced the...Ch. 39 - Prob. 79PQCh. 39 - Prob. 80PQCh. 39 - How much work is required to increase the speed of...Ch. 39 - Prob. 82PQCh. 39 - Prob. 83PQCh. 39 - Prob. 84PQCh. 39 - Prob. 85PQ
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- What does it tell us about factors on which air drag depends if it is proportional to speed squared?arrow_forwardWhat is the net charge on a sphere that has the following? x (a) 5.75 × 106 electrons and 8.49 × 106 protons 4.39e-13 What is the charge of an electron? What is the charge of a proton? C (b) 200 electrons and 109 protons 1.60e-10 What is the charge of an electron? What is the charge of a proton? Carrow_forwardA spider begins to spin a web by first hanging from a ceiling by his fine, silk fiber. He has a mass of 0.025 kg and a charge of 3.5 μC. A second spider with a charge of 4.2 μC rests in her own web exactly 2.1 m vertically below the first spider. (a) What is the magnitude of the electric field due to the charge on the second spider at the position of the first spider? 8.57e3 N/C (b) What is the tension in the silk fiber above the first spider? 0.125 How does the electric field relate to the force? How do you calculate the net force? Narrow_forward
- Point charges of 6.50 μC and -2.50 μC are placed 0.300 m apart. (Assume the negative charge is located to the right of the positive charge. Include the sign of the value in your answers.) (a) Where can a third charge be placed so that the net force on it is zero? 0.49 m to the right of the -2.50 μC charge (b) What if both charges are positive? 0.185 xm to the right of the 2.50 μC chargearrow_forwardc = ad Find the electric field at the location of q, in the figure below, given that q₁ = 9₁ = 9₁ = +4.60 nC, q=-1.00 nC, and the square is 20.0 cm on a side. (The +x axis is directed to the right.) magnitude direction N/C ° counterclockwise from the +x-axis 9a % 9 9barrow_forwardPlastic beads can often carry a small charge and therefore can generate electric fields. Three beads are oriented such that 92 is between q₁ and 93. The sum of the charge on 9₁ and 92 is 9₁ + 92 = −2.9 µС, and the net charge of the system of all three beads is zero. E field lines 93 92 What charge does each bead carry? 91 92 -1.45 What is the net charge of the system? What charges have to be equal? μC 2.9 ✓ What is the net charge of the system? What charges have to be equal? μC 93 2.9 μεarrow_forward
- A spider begins to spin a web by first hanging from a ceiling by his fine, silk fiber. He has a mass of 0.025 kg and a charge of 3.5 μC. A second spider with a charge of 4.2 μC rests in her own web exactly 2.1 m vertically below the first spider. (a) What is the magnitude of the electric field due to the charge on the second spider at the position of the first spider? 8.57e3 N/C (b) What is the tension in the silk fiber above the first spider? 0.275 How does the electric field relate to the force? How do you calculate the net force? Narrow_forwardPlastic beads can often carry a small charge and therefore can generate electric fields. Three beads are oriented such that 92 is between 91 system of all three beads is zero. E field lines 91 92 93 X What charge does each bead carry? 91 = 92 = ?2.9 0 μC × What is the net charge of the system? What charges have to be equal? μC 93 2.9 με and 93. The sum of the charge on 91 and 92 is 91 +92 = -2.9 μC, and the net charge of thearrow_forwardAn electron has an initial speed of 5.26 x 100 m/s in a uniform 5.73 x 105 N/C strength electric field. The field accelerates the electron in the direction opposite to its initial velocity. (a) What is the direction of the electric field? opposite direction to the electron's initial velocity same direction as the electron's initial velocity not enough information to decide × What is the direction of the force on the electron? How does it compare to the direction of the electric field, considering the sign of the electron's charge? (b) How far does the electron travel before coming to rest? 0.0781 × What kinematic equation is relevant here? How do you calculate the force due to the electric field? m (c) How long does it take the electron to come to rest? 5.27e8 What is the final velocity of the electron? s (d) What is the electron's speed when it returns to its starting point? 5.26e6 m/sarrow_forward
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