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Unformatted text preview: Chapter 2 2.1. Four 10nC positive charges are located in the z = 0 plane at the corners of a square 8cm on a side. A fifth 10nC positive charge is located at a point 8cm distant from the other charges. Calculate the magnitude of the total force on this fifth charge for = : Arrange the charges in the xy plane at locations (4,4), (4,4), (4,4), and (4,4). Then the fifth charge will be on the z axis at location z = 4 2, which puts it at 8cm distance from the other four. By symmetry, the force on the fifth charge will be zdirected, and will be four times the z component of force produced by each of the four other charges. F = 4 2 q 2 4 d 2 = 4 2 (10 8 ) 2 4 (8 . 85 10 12 )(0 . 08) 2 = 4 . 10 4 N 2.2. A charge Q 1 = 0 . 1 C is located at the origin, while Q 2 = 0 . 2 C is at A (0 . 8 , . 6 , 0). Find the locus of points in the z = 0 plane at which the x component of the force on a third positive charge is zero. To solve this problem, the z coordinate of the third charge is immaterial, so we can place it in the xy plane at coordinates ( x,y, 0). We take its magnitude to be Q 3 . The vector directed from the first charge to the third is R 13 = x a x + y a y ; the vector directed from the second charge to the third is R 23 = ( x . 8) a x + ( y + 0 . 6) a y . The force on the third charge is now F 3 = Q 3 4 Q 1 R 13  R 13  3 + Q 2 R 23  R 23  3 = Q 3 10 6 4 . 1( x a x + y a y ) ( x 2 + y 2 ) 1 . 5 + . 2[( x . 8) a x + ( y + 0 . 6) a y ] [( x . 8) 2 + ( y + 0 . 6) 2 ] 1 . 5 We desire the x component to be zero. Thus, 0 = . 1 x a x ( x 2 + y 2 ) 1 . 5 + . 2( x . 8) a x [( x . 8) 2 + ( y + 0 . 6) 2 ] 1 . 5 or x [( x . 8) 2 + ( y + 0 . 6) 2 ] 1 . 5 = 2(0 . 8 x )( x 2 + y 2 ) 1 . 5 2.3. Point charges of 50nC each are located at A (1 , , 0), B ( 1 , , 0), C (0 , 1 , 0), and D (0 , 1 , 0) in free space. Find the total force on the charge at A . The force will be: F = (50 10 9 ) 2 4 R CA  R CA  3 + R DA  R DA  3 + R BA  R BA  3 where R CA = a x a y , R DA = a x + a y , and R BA = 2 a x . The magnitudes are  R CA  =  R DA  = 2, and  R BA  = 2. Substituting these leads to F = (50 10 9 ) 2 4 1 2 2 + 1 2 2 + 2 8 a x = 21 . 5 a x N where distances are in meters. 15 2.4. Let Q 1 = 8 C be located at P 1 (2 , 5 , 8) while Q 2 = 5 C is at P 2 (6 , 15 , 8). Let = . a) Find F 2 , the force on Q 2 : This force will be F 2 = Q 1 Q 2 4 R 12  R 12  3 = (8 10 6 )( 5 10 6 ) 4 (4 a x + 10 a y ) (116) 1 . 5 = ( 1 . 15 a x 2 . 88 a y )mN b) Find the coordinates of P 3 if a charge Q 3 experiences a total force F 3 = 0 at P 3 : This force in general will be: F 3 = Q 3 4 Q 1 R 13  R 13  3 + Q 2 R 23  R 23  3 where R 13 = ( x 2) a x +( y 5) a y and R 23 = ( x 6) a x +( y 15) a y . Note, however, that all three charges must lie in a straight line, and the location of...
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This note was uploaded on 10/02/2011 for the course EE 1 taught by Professor Joshi during the Fall '08 term at UCLA.
 Fall '08
 Joshi
 Electrical Engineering

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