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# PHY7 - Howard(clh2528 – homework#7 – shubeita –(57885...

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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 μ C 9 μ C − 6 . 39 μ C P ˆ ı ˆ 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 μ C = 6 . 7 × 10 − 6 C , q 2 = 9 μ C = 9 × 10 − 6 C , q 3 = − 6 . 39 μ 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 μ C charge due to the 9 μ C charge? 1. F = k e (6 . 7 μ C) (9 μ C) (0 . 76 m) 2 2. F = k e (6 . 7 μ C) (9 μ C) (0 . 76 m) 2 tan 45 ◦ 3. F = k e (6 . 7 μ C) (9 μ C) (0 . 76 m) 2 cot 60 ◦ 4. F = k e (6 . 7 μ C) (9 μ C) (0 . 76 m) 2 tan 30 ◦ 5. F = k e (6 . 7 μ C) (9 μ C) (0 . 76 m) 2 tan 60 ◦ 6. F = k e (6 . 7 μ C) (9 μ C) (0 . 76 m) 2 cos 30 ◦ correct 7. F = k e (6 . 7 μ C) (9 μ C) (0 . 76 m) 2 sin 45 ◦ 8. F = k e (6 . 7 μ C) (9 μ C) (0 . 76 m) 2 cot 45 ◦ 9. F = k e (6 . 7 μ C) (9 μ C) (0 . 76 m) 2 cos 60 ◦ 10. F = k e (6 . 7 μ C) (9 μ C) (0 . 76 m) 2 cot 30 ◦ Explanation: F v = F cos α = k e q 1 q 2 r 2 cos α = k e (6 . 7 μ C) (9 μ 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 π ǫ r , where r is the distance between the charges, so, since k e = 1 4 π ǫ , 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...

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