PHY7 - Howard (clh2528) homework #7 shubeita (57885) 1 This...

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Unformatted text preview: Howard (clh2528) homework #7 shubeita (57885) 1 This print-out should have 16 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 (part 1 of 3) 10.0 points Consider three point charges at the vertices of an equilateral triangle. Let the potential be zero at infinity. . 7 6 m 60 6 . 7 C 9 C 6 . 39 C P What is the electrostatic potential at the point P at the center of the base of the equilateral triangle given in the dia- gram? The value of the Coulomb constant is 8 . 98755 10 9 N m 2 / C 2 . Correct answer: 1 . 5322 10 5 V. Explanation: Let : q 1 = 6 . 7 C = 6 . 7 10 6 C , q 2 = 9 C = 9 10 6 C , q 3 = 6 . 39 C = 6 . 39 10 6 C , a = 0 . 76 m , and k e = 8 . 98755 10 9 N m 2 / C 2 . The potential at P is given by V = k e summationdisplay i q i r i . The height h is h = radicalbigg a 2 parenleftBig a 2 parenrightBig 2 = 3 2 a . and q 1 > 0, q 2 > 0, and q 3 < 0. V P = k e parenleftbigg q 1 h + q 2 a/ 2 + q 3 a/ 2 parenrightbigg = 2 k e a parenleftbigg q 1 3 + q 2 + q 3 parenrightbigg = 2 ( 8 . 98755 10 9 N m 2 / C 2 ) . 76 m parenleftbigg 6 . 7 10 6 C 3 + 9 10 6 C 6 . 39 10 6 C ) = 1 . 5322 10 5 V . 002 (part 2 of 3) 10.0 points What is the vertical component of the electric force on the 6 . 7 C charge due to the 9 C charge? 1. F = k e (6 . 7 C) (9 C) (0 . 76 m) 2 2. F = k e (6 . 7 C) (9 C) (0 . 76 m) 2 tan 45 3. F = k e (6 . 7 C) (9 C) (0 . 76 m) 2 cot 60 4. F = k e (6 . 7 C) (9 C) (0 . 76 m) 2 tan 30 5. F = k e (6 . 7 C) (9 C) (0 . 76 m) 2 tan 60 6. F = k e (6 . 7 C) (9 C) (0 . 76 m) 2 cos 30 correct 7. F = k e (6 . 7 C) (9 C) (0 . 76 m) 2 sin 45 8. F = k e (6 . 7 C) (9 C) (0 . 76 m) 2 cot 45 9. F = k e (6 . 7 C) (9 C) (0 . 76 m) 2 cos 60 10. F = k e (6 . 7 C) (9 C) (0 . 76 m) 2 cot 30 Explanation: F v = F cos = k e q 1 q 2 r 2 cos = k e (6 . 7 C) (9 C) (0 . 76 m) 2 cos 30 003 (part 3 of 3) 10.0 points Find the total electrostatic energy of the sys- tem, again with the zero reference at infinity. Howard (clh2528) homework #7 shubeita (57885) 2 Correct answer: . 473301 J. Explanation: The total electrostatic energy of the sys- tem is the sum of the electrostatic energies between each pair of charges: U = U 12 + U 23 + U 31 The electrostatic energy between the charges q i and q j is given by U ij = q i q j 4 r , where r is the distance between the charges, so, since k e = 1 4 , U = k e a ( q 1 q 2 + q 2 q 3 + q 3 q 1 ) = ( 8 . 98755 10 9 N m 2 / C 2 ) (0 . 76 m) bracketleftBig (6 . 7 10 6 C) (9 10 6 C) + (9 10 6 C) ( 6 . 39 10 6 C) + ( 6 . 39 10 6 C) (6 . 7 10 6 C) bracketrightBig = . 473301 J ....
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PHY7 - Howard (clh2528) homework #7 shubeita (57885) 1 This...

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