Chap22_08 - Chapter 22 22.2 Sketch qualitatively the...

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Chapter 22 22.2 Sketch qualitatively the electric field lines both between and outside two concentric conducting spherical shells when uniform positive charge q 1 is on the inner shell and a uniform negative charge q 2 is on the outer. Consider the cases q 1 > q 2 , q 1 = q 2 , and q 1 < q 2 . q1> q2 q1= q2 q1< q2 22.5 An atom of plutonium-239 has a nuclear radius of 6.64 fm and atomic number Z=94. Assuming that the positive charge is distributed uniformly within the nucleus, what are the magnitude and direction of the electric field at the surface of the nucleus due to the positive charge. Outside a uniformly charged sphere, the field looks like that of a point charge at the center of the sphere. E = q 4 πε 0 r 2 = 94 1.6 × 10 19 C 4 0 (6.64 × 10 15 m ) 2 = 3.07 × 10 21 N / C ! 22.9 In Fig 22-32, the four particles form a square of edge length a = 5.00 cm and have charges q 1 = + 10 nC q 2 = 20 nC q 3 = + 20 nC q 4 = 10 nC . In unit-vector notation, what net electric field do the particle produce at the square’s center.
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q 1 q 2 q 3 q 4 E 1 E 2 E 3 E 4 We can write out each field in unit vector notation and then add. Here all the q’s are magnitudes since we have already included the directions by hand. r = 1 2 a 2 + a 2 = 2 a E 1 = q 1 4 πε 0 a 2 /2 cos45 ˆ i q 1 4 0 a 2 / 2 sin45 ˆ j = q 1 4 0 a 2 ˆ i q 1 2 4 0 a 2 ˆ j E 2 = q 2 4 0 a 2 / 2 cos45 ˆ i + q 2 4 0 a 2 / 2 sin45 ˆ j = q 2 4 0 a 2 ˆ i + q 2 4 0 a 2 ˆ j E 3 = q 3 4 0 a 2 / 2 ˆ i + q 3 4 0 a 2 / 2 sin45 ˆ j = q 3 4 0 a 2 ˆ i + q 3 4 0 a 2 ˆ j E 4 = q 4 4 0 a 2 / 2 cos45 ˆ i q 1 4 0 a 2 sin45 ˆ j = q 4 4 0 a 2 ˆ i q 4 4 0 a 2 ˆ j E = q 1 4 0 a 2 + q 2 4 0 a 2 q 3 2 4 0 a 2 q 4 4 0 a 2 ˆ i + q 1 4 0 a 2 + q 2 4 0 a 2 + q 3 4 0 a 2 q 4 4 0 a 2 ˆ j E = 4 0 a 2 q 1 + q 2 q 3 q 4 ( ) ˆ i + 4 0 a 2 q 1 + q 2 + q 3 q 4 ( ) ˆ j = 0 ˆ i + 1.017 × 10 5 ˆ j 22.10 In Fig. 22-33, what is the magnitude of the electric field at point P due to the four point charges shown? The fields due to the two +5q charges cancel exactly. We only need consider the fields due to the
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+3q and -12q charges. P +5q +3q -12q E 12 E 3 E = + 3 q 4 πε 0 d 2 + 12 q 4 0 (2 d ) 2 = 0! 22.13 In Fig. 22-36, three particles are fixed in place and have charges q 1 = q 2 = + e and q 3 = + 2 e . Distance a = 6.00 μ m . What are the (a)magnitude and (b) direction of the net electric field at point P due to the particles. q 1 q 2 q 3 E 1 E 2 E 3 The fields due to charges 1 and 2 cancel exactly. This leaves us only to calculate the field due to 3.
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r = 1 2 2 + a 2 = 2 a E 3 = q 3 4 πε 0 a 2 /2 = 159.8 N / C at 45 ° 22.15 Figure 22-38 shows a proton (pP on the central axis through a disk with a uniform charge density due to excess electrons . Three of those electrons are shown: electron e c at the disk center
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Chap22_08 - Chapter 22 22.2 Sketch qualitatively the...

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