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Unformatted text preview: Kapoor (mk9499) – oldhomework 09 – Turner – (60230) 1 This printout should have 12 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 2) 10.0 points Two spherical conductors (shown in the fig ure below) are separated by a distance large enough to make induction effects negligible. The spheres (with radii 15 cm and 9 cm) are connected by a thin conducting wire and are brought to the same potential of 775 V rela tive to 0 V at r = ∞ . 15 cm 9 cm Q 1 Q 2 Determine the capacitance of the sys tem. The Coulomb constant is 8 . 98755 × 10 9 N · m 2 / C 2 . Correct answer: 2 . 67036 × 10 − 11 F. Explanation: Let : k e = 8 . 98755 × 10 9 N · m 2 / C 2 , R 1 = 15 cm = 0 . 15 m , and R 2 = 9 cm = 0 . 09 m . R 1 R 2 Q 1 Q 2 Both spherical conductors are at the same potential V , so V = k e Q 1 R 1 = k e Q 2 R 2 . The charges on the spheres are then Q 1 = V R 1 k e and Q 2 = V R 2 k e . The capacitance of the system is C = Q 1 + Q 2 V = R 1 + R 2 k e = . 15 m + 0 . 09 m 8 . 98755 × 10 9 N · m 2 / C 2 = 2 . 67036 × 10 − 11 F . 002 (part 2 of 2) 10.0 points What is the charge ratio Q 1 Q 2 ? Correct answer: 1 . 66667. Explanation: From Eq. 2, Q ∝ R , so we have Q 1 Q 2 = R 1 R 2 = 15 cm 9 cm = 1 . 66667 . 003 10.0 points A parallel plate capacitor is connected to a battery. + Q Q d b b 2 d b b If we double the plate separation, 1. the potential difference is halved. 2. the capacitance is doubled. 3. the electric field is doubled. 4. None of these. 5. the charge on each plate is halved. cor rect Explanation: Kapoor (mk9499) – oldhomework 09 – Turner – (60230) 2 The capacitance of a parallel plate capaci tor is C = ǫ A d . Hence doubling d halves the capacitance, and Q = C V is also halved parenleftbigg C ′ = ǫ A 2 d = 1 2 ǫ A d = 1 2 C parenrightbigg . 004 (part 1 of 4) 10.0 points An airfilled capacitor consists of two parallel plates, each with an area of 7 cm 2 , separated by a distance 3 . 2 mm . A 24 V potential differ ence is applied to these plates. The permittivity of a vacuum is 8 . 85419 × 10 − 12 C 2 / N · m 2 . 1 pF is equal to 10 − 12 F . The magnitude of the electric field between the plates is 1. E = ( V d ) 2 . 2. E = parenleftbigg V d parenrightbigg 2 . 3. None of these 4. E = 1 ( V d ) 2 . 5. E = d V . 6. E = 1 V d . 7. E = parenleftbigg d V parenrightbigg 2 . 8. E = V d . correct 9. E = V d . Explanation: Since E is constant between the plates, V = integraldisplay vector E · d vector l = E d E = V d . 005 (part 2 of 4) 10.0 points The magnitude of the surface charge density on each plate is 1. σ = ǫ d V . 2. σ = ǫ V d 3. σ = ǫ V d . correct 4. None of these 5. σ = ǫ parenleftbigg d V parenrightbigg 2 ....
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This note was uploaded on 06/28/2009 for the course PHY 303L taught by Professor Turner during the Fall '08 term at University of Texas at Austin.
 Fall '08
 Turner
 Physics, Work

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