626_PartUniversity Physics Solution

626_PartUniversity Physics Solution - Electric Charge and...

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Electric Charge and Electric Field 21-19 The components of the two fields are shown in Figure 21.49g. 12 2 0 1 4 q EE r π == P 9 92 2 1 2 6.00 10 C (8.988 10 N m /C ) (0.250 m) E × 862.8 N/C = = Figure 21.49g 11 2 2 cos , cos xx θ =− =+ 0 xxx EEE =+= sin , sin yy ( )( ) 1 1 2 2 sin 2 862.8 N/C 0.800 1380 N/C yyy y EE E E E = = = 1380 N/C, in the -direction. E y EVALUATE: Point a is symmetrically placed between identical charges, so symmetry tells us the electric field must be zero. Point b is to the right of both charges and both electric fields are in the + x -direction and the resultant field is in this direction. At point c both fields have a downward component and the field of 2 q has a component to the right, so the net E G is in the 4th quadrant. At point d both fields have an upward component but by symmetry they have equal and opposite x -components so the net field is in the + y -direction. We can use this sort of reasoning to deduce the general direction of the net field before doing any calculations. 21.50. IDENTIFY: Apply Eq.(21.7) to calculate the field due to each charge and then calculate the vector sum of those fields. SET UP: The fields due to 1 q and to 2 q are sketched in Figure 21.50. EXECUTE: 9 2 2 0 1 (6.00 10 C) ˆˆ () 1 5 0 N / C 4( 0 . 6 m ) × = Ei i G P . 9 1 22 0 1 ˆ ˆ (4.00 10 C) (0.600) (0.800) (21.6 28.8 )N C 1 . 0 0 m ) ( 1 . 0 0 m ) ⎛⎞ + = ⎜⎟ ⎝⎠ j i + j G P . ( 128.4 N/C) (28.8 N/C) E=E +E i+ j GGG . (128.4 N/C) (28.8 N/C) 131.6 N/C E = at 1 28.8 tan 12.6 128.4 ° above the x axis and therefore 196.2 ° counterclockwise from the + x axis. EVALUATE: 1 E G is directed toward 1 q because 1 q is negative and 2 E G is directed away from 2 q because 2 q is positive. Figure 21.50 21.51. IDENTIFY: The resultant electric field is the vector sum of the field of and of . qq G G SET UP: The placement of the charges is shown in Figure 21.51a. Figure 21.51a
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21-20 Chapter 21 EXECUTE: (a) The directions of the two fields are shown in Figure 21.51b. 1 12 2 01 1 4 q EE r π == P 9 92 2 1 2 6.00 10 C (8.988 10 N m /C ) (0.150 m) E × 2397 N/C = = Figure 21.51b 11 2 x2 2397 N/C, 0 2397 N/C, 0 xy y =− = = 2( 2397 N/C) 4790 N/C xxx EE E =+= = 0 yyy EEE The resultant electric field at point a in the sketch has magnitude 4790 N/C and is in the -direction. x (b) The directions of the two fields are shown in Figure 21.51c.
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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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626_PartUniversity Physics Solution - Electric Charge and...

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