Chap23_04 - Chapter 23 23.4 What is the magnitude of a...

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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 πε 0 r 2 q = 4 0 r 2 E = 1.112 × 10 10 C 23.6 Two particles with equal charge magnitudes 2.0 × 10 7 C but opposite signes are held 15cm aprat what are the magnitude and direction of the E at the point midway between the charges? E - E + +q -q The electric field due to each charge points to the right in the picture above. E = E + + E = q 4 0 r 2 + q 4 0 r 2 = 2.0 × 10 7 4 0 (0.075) 2 + 2.0 × 10 7 4 0 (0.075) 2 = 6.4 × 10 5
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23.8 In Fig. 23-27, two fixed point charges q 1 = + 1.0 × 10 6 and q 2 = + 3.0 × 10 6 C are separated by a distance d=10cm. Plot their net electric field as a funciton of x for poth positive and negative values of x, taking E to be positive when the vector E point s to the right and negative when E points to the left. The graph shows a plot of the equation: y = 1 × 10 6 4 πε 0 x 2 x x 2 ˆ i + 3 × 10 6 4 0 ( x 0.1) 2 ( x x 2 ˆ i
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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. P +5q +3q -12q E 12 E 3 E = + 3 q 4 πε 0 d 2 + 12 q 4 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 +2q E 2 E = + 2 q 4 0 ( a 2 /2) at 45 °
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23.16. 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 πε 0 r + 2 ˆ r + q 4 0 r 2 ˆ r r r + = ( r 0) ˆ i + (0 d 2 ) ˆ j 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 πε 0 r + 2 ˆ r + q 4 0 r 2 ˆ r = q 4 0 ( r 2 + ( d 2 ) 2 ) r ˆ i d 2 ˆ j r 2 + ( d 2 ) 2 q 4 0 ( r 2 + ( d 2 ) 2 ) r ˆ i + d 2 ˆ j 2 + ( d 2 ) 2 = qd 4 0 ( r 2 + ( d 2 ) 2 ) 3/ 2 ˆ j = p 4 0 r 3 (1 + ( d 2 r ) 2 ) 3/2 ˆ j r p 4 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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This note was uploaded on 05/22/2008 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.

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Chap23_04 - Chapter 23 23.4 What is the magnitude of a...

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