6 - kavo (ak22862) – HW06 – Janow – (12121) 1 This...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: kavo (ak22862) – HW06 – Janow – (12121) 1 This print-out should have 17 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 conductors having net charges of 14 . 9 μ C and- 14 . 9 μ C have a potential difference of 8 . 4 V. Determine the capacitance of the system. Correct answer: 1 . 77381 × 10 − 6 F. Explanation: Let : Q = 14 . 9 μ C = 1 . 49 × 10 − 5 C and V = 8 . 4 V . The capacitance is C = Q V = 1 . 49 × 10 − 5 C 8 . 4 V = 1 . 77381 × 10 − 6 F . 002 (part 2 of 2) 10.0 points Determine the potential difference between the two conductors if the charges on each are increased to 54 μ C magnitude. Correct answer: 30 . 443 V. Explanation: Let : Q 2 = 54 μ C = 5 . 4 × 10 − 5 C . The new potential difference between the two conductors is V 2 = Q 2 C = 5 . 4 × 10 − 5 C 1 . 77381 × 10 − 6 F = 30 . 443 V . 003 (part 1 of 4) 10.0 points An air-filled capacitor consists of two parallel plates, each with an area of 9 . 2 cm 2 , sepa- rated by a distance 2 . 5 mm . A 16 V potential difference 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 = parenleftbigg d V parenrightbigg 2 . 2. E = 1 V d . 3. E = V d . 4. E = d V . 5. E = 1 ( V d ) 2 . 6. E = parenleftbigg V d parenrightbigg 2 . 7. E = V d . correct 8. E = ( V d ) 2 . 9. None of these Explanation: Since E is constant between the plates, V = integraldisplay vector E · d vector l = E d E = V d . 004 (part 2 of 4) 10.0 points The magnitude of the surface charge density on each plate is 1. σ = ǫ V d . correct 2. σ = ǫ V d . 3. None of these kavo (ak22862) – HW06 – Janow – (12121) 2 4. σ = ǫ ( V d ) 2 . 5. σ = ǫ parenleftbigg V d parenrightbigg 2 . 6. σ = ǫ d V . 7. σ = ǫ parenleftbigg d V parenrightbigg 2 . 8. σ = ǫ V d 9. σ = ǫ ( 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 E = ǫ V d Alternatively, we could just recall this result for an infinite conducting plate (meaning we neglect edge effects) and apply it....
View Full Document

This note was uploaded on 09/16/2009 for the course PHYS 121 taught by Professor Opyrchal during the Fall '08 term at NJIT.

Page1 / 6

6 - kavo (ak22862) – HW06 – Janow – (12121) 1 This...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online