EBK PHYSICS FOR SCIENTISTS & ENGINEERS
5th Edition
ISBN: 9780134296074
Author: GIANCOLI
Publisher: VST
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Chapter 22, Problem 36P
(II) A thin cylindrical shell of radius R1 = 6.5 cm is surrounded by a second cylindrical shell of radius R2 = 9.0 cm, as in Fig. 22–35. Both cylinders are 5.0 m long and the inner one carries a total charge Q1 = −0.88 μC and the outer one Q2 = +1.56 μC. For points far from the ends of the cylinders, determine the electric field at a radial distance r from the central axis of (a) 3.0 cm, (b) 7.0 cm, and (c) 12.0 cm.
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(II) Two point charges, Q₁ = -25 μC and Q2 = +45 μC,
are separated by a distance of 12 cm. The electric field at the
point P (see Fig. 21-58) is zero. How far from Q₁ is P?
21
-25 μC
FIGURE 21-58
Problem 36. P
X
12 cm
22
+45 μC
(II) Determine the electric field É at the origin 0 in
Fig. 16–58 due to the two charges
at A and B.
y
|+26 µC
A
8.0 cm
-26 µC
B
8.0 cm
FIGURE 16-58
8.0 cm
Problem 33.
(II) The electric field between two parallel square metal plates is 130 N/C. The plates are 0.85 m on a side and are separated by 3.0 cm. What is the charge on each plate (assume equal and opposite)? Neglect edge effects
Chapter 22 Solutions
EBK PHYSICS FOR SCIENTISTS & ENGINEERS
Ch. 22.1 - Which of the following would cause a change in the...Ch. 22.2 - A point charge Q is at the center of a spherical...Ch. 22.2 - Three 2.95 C charges are in a small box. What is...Ch. 22.3 - Prob. 1EECh. 22 - If the electric flux through a closed surface is...Ch. 22 - Is the electric field E in Gausss law....Ch. 22 - What can you say about the flux through a closed...Ch. 22 - The electric field E is zero at all points on a...Ch. 22 - Define gravitational flux in analogy to electric...Ch. 22 - Would Gausss law be helpful in determining the...
Ch. 22 - A spherical basketball (a nonconductor) is given a...Ch. 22 - In Example 226, it may seem that the electric...Ch. 22 - Suppose the line of charge in Example 226 extended...Ch. 22 - A point charge Q is surrounded by a spherical...Ch. 22 - A solid conductor carries a net positive charge Q....Ch. 22 - A point charge q is placed at the center of the...Ch. 22 - A small charged ball is inserted into a balloon....Ch. 22 - Prob. 1MCQCh. 22 - Prob. 2MCQCh. 22 - Prob. 3MCQCh. 22 - Prob. 4MCQCh. 22 - Prob. 5MCQCh. 22 - Prob. 6MCQCh. 22 - Prob. 7MCQCh. 22 - Prob. 8MCQCh. 22 - Prob. 9MCQCh. 22 - Prob. 10MCQCh. 22 - Prob. 1PCh. 22 - (I) The Earth possesses an electric field of...Ch. 22 - (II) A cube of side l is placed in a uniform field...Ch. 22 - (II) A uniform field E is parallel to the axis of...Ch. 22 - (I) The total electric flux from a cubical box...Ch. 22 - (I) Figure 2226 shows five closed surfaces that...Ch. 22 - (II) In Fig. 2227, two objects, O1 and O2, have...Ch. 22 - (II) A ring of charge with uniform charge density...Ch. 22 - (II) In a certain region of space, the electric...Ch. 22 - (II) A point charge Q is placed at the center of a...Ch. 22 - Prob. 11PCh. 22 - (I) Draw the electric field lines around a...Ch. 22 - Prob. 13PCh. 22 - (I) Starting from the result of Example 223, show...Ch. 22 - Prob. 15PCh. 22 - (I) A metal globe has l.50 mC of charge put on it...Ch. 22 - Prob. 17PCh. 22 - (II) A solid metal sphere of radius 3.00 m carries...Ch. 22 - (II) A 15.0-cm-diameter nonconducting sphere...Ch. 22 - (II) A flat square sheet of thin aluminum foil,...Ch. 22 - (II) A spherical cavity of radius 4.50 cm is at...Ch. 22 - Prob. 22PCh. 22 - Prob. 23PCh. 22 - (II) Two large, flat metal plates are separated by...Ch. 22 - (II) Suppose the two conducting plates in Problem...Ch. 22 - Prob. 26PCh. 22 - (II) Two thin concentric spherical shells of radii...Ch. 22 - (II) A spherical rubber balloon carries a total...Ch. 22 - (II) Suppose the nonconducting sphere of Example...Ch. 22 - (II) Suppose in Fig. 2232, Problem 29, there is...Ch. 22 - (II) Suppose the thick spherical shell of Problem...Ch. 22 - (II) Suppose that at the center of the cavity...Ch. 22 - (II) A long cylindrical shell of radius R0 and...Ch. 22 - (II) A very long solid nonconducting cylinder of...Ch. 22 - (II) A thin cylindrical shell of radius R1 is...Ch. 22 - (II) A thin cylindrical shell of radius R1 = 6.5...Ch. 22 - (II) (a) If an electron (m = 9.1 1031 kg) escaped...Ch. 22 - (II) A very long solid nonconducting cylinder of...Ch. 22 - (II) A nonconducting sphere of radius r0 is...Ch. 22 - (II) A very long solid nonconducting cylinder of...Ch. 22 - (II) A flat ring (inner radius R0, outer radius...Ch. 22 - (II) An uncharged solid conducting sphere of...Ch. 22 - (III) A very large (i.e., assume infinite) flat...Ch. 22 - (III) Suppose the density of charge between r1 and...Ch. 22 - (III) Suppose two thin flat plates measure 1.0 m ...Ch. 22 - (III) A flat slab of nonconducting material (Fig....Ch. 22 - (III) A flat slab of nonconducting material has...Ch. 22 - (III) An extremely long, solid nonconducting...Ch. 22 - (III) Charge is distributed within a solid sphere...Ch. 22 - Prob. 50GPCh. 22 - Prob. 51GPCh. 22 - The Earth is surrounded by an electric field,...Ch. 22 - Prob. 53GPCh. 22 - Prob. 54GPCh. 22 - Prob. 55GPCh. 22 - Prob. 57GPCh. 22 - Prob. 58GPCh. 22 - Prob. 59GPCh. 22 - Prob. 60GPCh. 22 - Prob. 61GPCh. 22 - Prob. 62GPCh. 22 - Prob. 63GPCh. 22 - Prob. 64GPCh. 22 - Prob. 65GPCh. 22 - Prob. 66GP
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- (c) As shown in Figure 3, there are 2 non-conducting rings each with uniform charge q1 and q2. Both rings have the same radius R. The separation distance between the rings is d = 4.0 R. Given q1 = 10.0 nC, q2 = -20.0 nC, R= 0.50 m, d= 1.50 m. Calculate the net electric field at point P. Ring 1 Ring 2 12 P R R -R→| d Figure 3arrow_forward(II) Two point charges, Q1 = -32 µC and Q2 = +45 µC, are separated by a distance of 12 cm. The electric field at the point P (see Fig. 16–57) is zero. How far from Qj is P? Q1 Q2 12 cm P -32 μC +45 µC FIGURE 16-57 Problem 32.arrow_forwardA very thin filament of uniform linear charge density "A" is located on the x-axis from x=0 to x=a. Prove that the components of the electric field at a point P on the y-axis, located at the distance "y" from the origin are:Ex = -k^(1/y-1/√/y² + a²) i, Ey = kha/y√/y² + a²)]arrow_forward
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