Ch16-h3-solutions - pan (zp695) Ch16-h3 chiu (56565) This...

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pan (zp695) – Ch16-h3 – chiu – (56565) 1 This print-out should have 19 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 (part 1 of 4) 10.0 points A solid metal ball oF radius 1 . 5 cm bearing acha rg eo F - 18 nC is located near a solid plastic ball oF radius 2 cm bearing a uniFormly distributed charge oF 5 nC on its outer surFace. The distance between the centers oF the balls is 11 cm. Given that the value oF k is 9 × 10 9 N · m 2 / C 2 . - 18 nC 5nC Metal Plastic Which one oF the Following best represents the charge distributions on the two balls? 1. - 18 nC Metal Plastic - - - - - - - - - - - - + + + + + + + + 2. - 18 nC Metal Plastic -- - - - - - - ++ + + + + + + 3. - 18 nC Metal Plastic - - - - - - - - - - - - + + + + + + correct 4. No static charge distribution will satisFy this setup–the charges will be moving For all times. 5. - 18 nC Metal Plastic - - - - - - + + + + + + + + Explanation: Plastic is an insulator, so the initially uni- Formly distributed positive charges will not be able to move, even when brought near the metal ball. Metal, however, is a conductor, and since the positive charges on the plastic ball will attract the negative charges on the metal ball, this will cause a net migration oF negative charge to the right hand side oF the metal ball. ThereFore, the actual charge distribution wil most closely mirror this one: - 18 nC Metal Plastic - - - - - - - - - - - - + + + + + + 002 (part 2 of 4) 10.0 points What is the magnitude oF the electric feld at the center oF the metal ball due only to the charges on the plastic ball? Correct answer: 3719 . 01 N / C.
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pan (zp695) – Ch16-h3 – chiu – (56565) 2 Explanation: Gauss’s law tells us that we can treat the feld outside oF a spherically symmetric charge distribution precisely as iF it were a point charge with all oF the charge concentrated at the center oF the distribution. Since the center oF the plastic ball is 11 cm away From the center oF the metal ball, we fnd that the net contribution due to the plastic ball is E = k q plastic d 2 = (9 × 10 9 N · m 2 / C 2 )(5 nC) (11 cm) 2 × 1 × 10 - 9 C / nC (0 . 01 m / cm) 2 =3719 . 01 N / C 003 (part 3 of 4) 10.0 points What is the net electric feld at the center oF the metal ball? Correct answer: 0 N / C. Explanation: The feld inside oF a conductor in a static confguration must be zero, since iF the feld were not zero, then the net feld would move charges in the conductor, creating a non-static charge distribution. 004 (part 4 of 4) 10.0 points What is the magnitude oF the electric feld at the center oF the metal ball due only to charges on the surFace oF the metal ball? Correct answer: 3719 . 01 N / C. Explanation: We know that the feld is zero inside oF a conductor. ±urthermore, we know by the principle oF superposition that the net feld at the center oF the ball is equal to the contribution due to the plastic ball plus the contribution due to the metal ball. Since the net feld is zero, the charges on the metal ball must precisely cancel the charges on the plastic ball. This is only possible iF the two contributions have equal magnitudes and opposite directions.
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Ch16-h3-solutions - pan (zp695) Ch16-h3 chiu (56565) This...

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