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chapter2 hayt

# chapter2 hayt - Chapter 2 2.1 Four 10nC positive charges...

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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 = 0 : 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 z -directed, and will be four times the z component of force produced by each of the four other charges. F = 4 2 × q 2 4 π 0 d 2 = 4 2 × (10 8 ) 2 4 π (8 . 85 × 10 12 )(0 . 08) 2 = 4 . 0 × 10 4 N 2.2. A charge Q 1 = 0 . 1 µ C is located at the origin, while Q 2 = 0 . 2 µ C is at A (0 . 8 , 0 . 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 0 . 8) a x + ( y + 0 . 6) a y . The force on the third charge is now F 3 = Q 3 4 π 0 Q 1 R 13 | R 13 | 3 + Q 2 R 23 | R 23 | 3 = Q 3 × 10 6 4 π 0 0 . 1( x a x + y a y ) ( x 2 + y 2 ) 1 . 5 + 0 . 2[( x 0 . 8) a x + ( y + 0 . 6) a y ] [( x 0 . 8) 2 + ( y + 0 . 6) 2 ] 1 . 5 We desire the x component to be zero. Thus, 0 = 0 . 1 x a x ( x 2 + y 2 ) 1 . 5 + 0 . 2( x 0 . 8) a x [( x 0 . 8) 2 + ( y + 0 . 6) 2 ] 1 . 5 or x [( x 0 . 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 , 0), B ( 1 , 0 , 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 π 0 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 π 0 1 2 2 + 1 2 2 + 2 8 a x = 21 . 5 a x µ N where distances are in meters. 15

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2.4. Let Q 1 = 8 µ C be located at P 1 (2 , 5 , 8) while Q 2 = 5 µ C is at P 2 (6 , 15 , 8). Let = 0 . a) Find F 2 , the force on Q 2 : This force will be F 2 = Q 1 Q 2 4 π 0 R 12 | R 12 | 3 = (8 × 10 6 )( 5 × 10 6 ) 4 π 0 (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 π 0 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 π 0 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 = 0 which reduces to 8( x 6) 2
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chapter2 hayt - Chapter 2 2.1 Four 10nC positive charges...

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