621_PartUniversity Physics Solution

621_PartUniversity Physics Solution - 21-14 Chapter 21...

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21-14 Chapter 21 EVALUATE: The magnitude of the field approaches infinity at the location of one of the point charges. Figure 21.43 21.44. IDENTIFY: For a point charge, 2 q E k r = . For the net electric field to be zero, 1 E G and 2 E G must have equal magnitudes and opposite directions. SET UP: Let 1 0.500 nC q =+ and 2 8.00 nC. q E G is toward a negative charge and away from a positive charge. EXECUTE: The two charges and the directions of their electric fields in three regions are shown in Figure 21.44. Only in region II are the two electric fields in opposite directions. Consider a point a distance x from 1 q so a distance 1.20 m x from 2 q . 12 E E = gives 22 0.500 nC 8.00 nC (1.20 ) kk x x = . 16 (1.20 m ) x x =− . 4 (1.20 m ) x x and 0.24 m x = is the positive solution. The electric field is zero at a point between the two charges, 0.24 m from the 0.500 nC charge and 0.96 m from the 8.00 nC charge. EVALUATE: There is only one point along the line connecting the two charges where the net electric field is zero. This point is closer to the charge that has the smaller magnitude. Figure 21.44 21.45. IDENTIFY: Eq.(21.7) gives the electric field of each point charge. Use the principle of superposition and add the electric field vectors. In part (b) use Eq.(21.3) to calculate the force, using the electric field calculated in part (a). (a) SET UP: The placement of charges is sketched in Figure 21.45a. Figure 21.45a The electric field of a point charge is directed away from the point charge if the charge is positive and toward the point charge if the charge is negative. The magnitude of the electric field is 2 0 1 , 4 q E r π = P where r is the distance between the point where the field is calculated and the point charge. (i) At point a the fields 11 2 2 of and of qq EE GG are directed as shown in Figure 21.45b. Figure 21.45b
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Electric Charge and Electric Field 21-15 EXECUTE: 9 1 92 2 1 22 01 12 . 0 0 1 0 C (8.988 10 N m /C ) 449.4 N/C 4 (0.200 m) q E r π × == × = P 9 2 2 2 02 15 . 0 0 1 0 C 124.8 N/C 4 (0.600 m) q E r × × = P 11 449.4 N/C, 0 xy EE 124.8 N/C, 0 449.4 N/C 124.8 N/C 574.2 N/C xxx EE E =+= + + = + 0 yyy EEE The resultant field at point a has magnitude 574 N/C and is in the +
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This note was uploaded on 07/16/2011 for the course PHY 2053 taught by Professor Buchler during the Spring '06 term at University of Florida.

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621_PartUniversity Physics Solution - 21-14 Chapter 21...

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