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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 )( . 08 ) 2 = 4 . 10 4 N 2.2. A charge Q 1 = . 1 C is located at the origin, while Q 2 = . 2 C is at A( . 8 , . 6 , ) . 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, ) . 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 + . 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 + . 6 ) a y ] [ (x . 8 ) 2 + (y + . 6 ) 2 ] 1 . 5 We desire the x component to be zero. Thus, = . 1 x a x (x 2 + y 2 ) 1 . 5 + . 2 (x . 8 ) a x [ (x . 8 ) 2 + (y + . 6 ) 2 ] 1 . 5 or x [ (x . 8 ) 2 + (y + . 6 ) 2 ] 1 . 5 = 2 ( . 8 x)(x 2 + y 2 ) 1 . 5 2.3. Point charges of 50nC each are located at A( 1 , , ) , B( 1 , , ) , C( , 1 , ) , and D( , 1 , ) 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. 14 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 Q 3 will be along the vector R 12 extended past Q 2 . The slope of this vector is ( 15 5 )/( 6 2 ) = 2 . 5. Therefore, we look for P 3 at coordinates (x, 2 . 5 x, 8 ) . With this restriction, the force becomes: F 3 = Q 3 4 , 8[ (x 2 ) a x + 2 . 5 (x 2 ) a y ] [ (x 2 ) 2 + ( 2 . 5 ) 2 (x 2 ) 2 ] 1 . 5 5[ (x 6 ) a x + 2 . 5 (x 6 ) a y ] [ (x 6 ) 2 + ( 2 . 5 ) 2 (x 6 ) 2 ] 1 . 5 where we require the term in large brackets to be zero. This leads to 8 (x 2 ) [ (( 2 . 5 ) 2 + 1 )(x 6 ) 2 ] 1 . 5 5 (x 6 ) [ (( 2 . 5 ) 2 + 1 )(x 2 ) 2 ] 1 . 5 = which reduces to 8 (x 6 ) 2...
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This note was uploaded on 06/10/2008 for the course EE 325 taught by Professor Brown during the Spring '08 term at University of Texas at Austin.
 Spring '08
 Brown
 Electromagnet

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