solution35

solution35 - cherubin(mcc2645 – Homework 1 Electrostatics...

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Unformatted text preview: cherubin (mcc2645) – Homework 1 Electrostatics – Catala – (12054) 1 This print-out should have 22 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A charge of +2 . 20 × 10 − 9 C is placed at the origin, and another charge of +5 . 90 × 10 − 9 C is placed at x = 2 . 5 m. Find the point (coordinate) between these two charges where a charge of +3 . 00 × 10 − 9 C should be placed so that the net electric force on it is zero. The Coulomb constant is 8 . 98755 × 10 9 N · m 2 / C 2 . Correct answer: 0 . 947822 m. Explanation: Let : q 1 = 2 . 20 × 10 − 9 C at the origin , q 2 = 5 . 90 × 10 − 9 C , q 3 = 3 . 00 × 10 − 9 C , and x 2 = 2 . 5 m . q 3 is between q 1 and q 2 , so if r 1 , 3 = P , then r 2 , 3 = x 2 − P . F electric = k C q 1 q 2 r 2 The net electric force is zero, so F 1 , 3 = F 2 , 3 k C q 1 q 3 r 2 1 , 3 = k C q 2 q 3 r 2 2 , 3 q 1 r 2 1 , 3 = q 2 r 2 2 , 3 q 1 P 2 = q 2 ( x 2 − P ) 2 ( x 2 − P ) 2 P 2 = q 2 q 1 x 2 − P P = radicalbigg q 2 q 1 x 2 − P = P radicalbigg q 2 q 1 P parenleftbigg 1 + radicalbigg q 2 q 1 parenrightbigg = x 2 P = x 2 1 + radicalbigg q 2 q 1 = 2 . 5 m 1 + radicalbigg 5 . 9 × 10 − 9 C 2 . 2 × 10 − 9 C = . 947822 m . 002 10.0 points Find the magnitude of the electric field at a point midway between two charges +33 . 5 × 10 − 9 C and +78 . 3 × 10 − 9 C separated by a distance of 15.0 cm. The value of the Coulomb constant is 8 . 99 × 10 9 N · m 2 / C 2 . Correct answer: 71600 . 4 N / C. Explanation: Let : q 1 = 33 . 5 × 10 − 9 C , q 2 = 78 . 3 × 10 − 9 C , x = 15 . 0 cm , and k C = 8 . 99 × 10 9 N · m 2 / C 2 . r 1 = r 2 = x 2 = 15 cm 2 · parenleftbigg 1 m 100 cm parenrightbigg = 0 . 075 m The electric field is E = k C q r 2 . The magnitudes of the electric fields are E 1 = 8 . 99 × 10 9 N · m 2 / C 2 × 3 . 35 × 10 − 8 C (0 . 075 m) 2 = 53540 . 4 N / C , and E 2 = 8 . 99 × 10 9 N · m 2 / C 2 × 7 . 83 × 10 − 8 C (0 . 075 m) 2 = 1 . 25141 × 10 5 N / C . cherubin (mcc2645) – Homework 1 Electrostatics – Catala – (12054) 2 If the 33 . 5 × 10 − 9 C charge is the left one and the positive direction is to the right, E net = E 1 − E 2 = 53540 . 4 N / C − 1 . 25141 × 10 5 N / C = − 71600 . 4 N / C bardbl vector E net bardbl = 71600 . 4 N / C . The electric field at midpoint between the charges +33 . 5 × 10 − 9 C and +78 . 3 × 10 − 9 C is 71600 . 4 N / C toward the +33 . 5 × 10 − 9 C charge. 003 (part 1 of 4) 10.0 points A small cork with an excess charge of +9 . μ C is placed 0 . 24 m from another cork, which carries a charge of − 3 . 2 μ C. What is the magnitude of the electric force between the corks? The Coulomb constant is 8 . 98755 × 10 9 N · m 2 / C 2 ....
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solution35 - cherubin(mcc2645 – Homework 1 Electrostatics...

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