hw2 - PHY108 FALL 08 Solutions for HW2(Due on 9/17) 6. With...

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PHY108 FALL 08 Solutions for HW2(Due on 9/17) 6. With x 1 = 6.00 cm and x 2 = 21.00 cm, the point midway between the two charges is located at x = 13.5 cm. The values of the charge are q 1 = – q 2 = – 2.00 10 –7 C, and the magnitudes and directions of the individual fields are given by: 5 1 1 2 01 5 2 2 2 || ˆˆ i (3.196 10 N C)i 4 ( ) i (3.196 10 N C)i 4 ( ) q E xx q E Thus, the net electric field is 5 net 1 2 ˆ (6.39 10 N C)i E E E 7. At points between the charges, the individual electric fields are in the same direction and do not cancel. Since charge q 2 = 4.00 q 1 located at x 2 = 70 cm has a greater magnitude than q 1 = 2.1 10 8 C located at x 1 = 20 cm, a point of zero field must be closer to q 1 than to q 2 . It must be to the left of q 1 . Let x be the coordinate of P , the point where the field vanishes. Then, the total electric field at P is given by 21 2 2 02 1 | | | | 1 4 ( ) qq E . If the field is to vanish, then 2 2 1 2 2 22 2 11 | | | | | | ( ) . ( ) | | q q q x x x x q x x x x Taking the square root of both sides, noting that | q 2 |/| q 1 | = 4, we obtain 70 2.0 20 x x . Choosing –2.0 for consistency, the value of x is found to be x = 30 cm.
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11. The x component of the electric field at the center of the square is given by 3 1 2 4 2 2 2 2 0 1 2 3 4 2 0 || | | | | | | 1 cos45 4 ( / 2) ( / 2) ( / 2) ( / 2) 1 1 1 | | | | | | | | 4 / 2 2 0.
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This note was uploaded on 03/26/2010 for the course PHY 108 taught by Professor Iashvili during the Spring '08 term at SUNY Buffalo.

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hw2 - PHY108 FALL 08 Solutions for HW2(Due on 9/17) 6. With...

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