Chap23_03

# Chap23_03 - .4 What is the magnitude of a point charge that...

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Chapter 23 23.4 What is the magnitude of a point charge that would create and electric field of 1.00N/C at points 1.00m away. E = q 4 p e 0 r 2 q = 4 p e 0 r 2 E = 1.112 10 - 10 C 23.7 An atom of plutonium-239 has a nuclear radius of 6.64fm 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 p e 0 r 2 = 94 1.6 10 - 19 C 4 p e 0 (6.64 10 - 15 m ) 2 = 3.07 10 21 N / C ! 23.11 In Fig. 23-28, 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 +3q and -12q charges.

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P +5q +5q +3q -12q E 12 E 3 E = + 3 q 4 p e 0 d 2 + - 12 q 4 pe 0 (2 d ) 2 = 0! 23.12. Calculate the direction and magnitude of the electric field at the point p. The fields due to the +q cancel exactly, We only need compute the field due to the +2q. P +q +q +2q E 2 E = + 2 q 4 p e 0 ( a 2 /2) at 45 °
16. Find the magnitude and direction of the electric field at point P due to the electric dipole. You may assume that r>>d +q -q (0,d/2) (0,-d/2) (r,0) We begin by writing the exact field. r E = r E + + r E - = q 4 p e 0 r + 2 ˆ r + - q 4 p e 0 r - 2 ˆ r - r r + = ( r - 0) ˆ i + (0 - d 2 ) ˆ j r + = r 2 + ( d 2 ) 2 ˆ r + = r r + r + = r ˆ i - d 2 ˆ j r 2 + ( d 2 ) 2 r r - = ( r - 0) ˆ i + (0 - ( - d 2 )) ˆ j r - = r 2 + ( d 2 ) 2 ˆ r - = r r - r - = r ˆ i + d 2 ˆ j r 2 + ( d 2 ) 2 We can substitute in. In the last step, since d/2r is very small we ignore this term. Remember that the direction of p is in the +j direction in this problem (from - to + charge).

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r E = r E + + r E - = q 4 p e 0 r + 2 ˆ r + - q 4 p e 0 r - 2 ˆ r - = q 4 p e 0 ( r 2 + ( d 2 ) 2 ) r ˆ i - d 2 ˆ j r 2 + ( d 2 ) 2 - q 4 p e 0 ( r 2 + ( d 2 ) 2 ) r ˆ i + d 2 ˆ j r 2 + ( d 2 ) 2 = - qd 4 p e 0 ( r 2 + ( d 2 ) 2 ) 3/ 2 ˆ j = - p 4 p e 0 r 3 (1 + ( d 2 r ) 2 ) 3/2 ˆ j - r p 4 p e 0 r 3 23.18 Figure 23-33 shows two parallel conducting rings with their central axes along a common line . Ring 1 has uniform charge q 1 and radius R; ring 2 has uniform charge q 2 and the same radius R. The rings are separated by a distance 3R. The net electric field at P on the common line at a distance R from ring 1 is zero. What is the ratio q 1 / q 2 ?
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