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Homework Old Combined

# Homework Old Combined - Kapoor(mk9499 oldhomework 09...

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Kapoor (mk9499) – oldhomework 09 – Turner – (60230) 1 This print-out should have 12 questions. Multiple-choice 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 = 0 . 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 2 d 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:

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Kapoor (mk9499) – oldhomework 09 – Turner – (60230) 2 The capacitance of a parallel plate capaci- tor is C = ǫ 0 A d . Hence doubling d halves the capacitance, and Q = C V is also halved parenleftbigg C = ǫ 0 A 2 d = 1 2 ǫ 0 A d = 1 2 C parenrightbigg . 004 (part 1 of 4) 10.0 points An air-filled 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 0 d V . 2. σ = ǫ 0 V d 3. σ = ǫ 0 V d . correct 4. None of these 5. σ = ǫ 0 parenleftbigg d V parenrightbigg 2 . 6. σ = ǫ 0 parenleftbigg V d parenrightbigg 2 . 7. σ = ǫ 0 V d . 8. σ = ǫ 0 ( V d ) 2 . 9. σ = ǫ 0 ( V d ) 2 . Explanation: Use Gauss’s Law. We find that a pillbox of cross section S which sticks through the sur- face on one of the plates encloses charge σ S. The flux through the pillbox is only through the top, so the total flux is E S. Gauss’ Law gives σ = ǫ 0 E = ǫ 0 V d
Kapoor (mk9499) – oldhomework 09 – Turner – (60230) 3 Alternatively, we could just recall this result for an infinite conducting plate (meaning we neglect edge effects) and apply it.

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