This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: lew (dl9564) – hw 5 – Opyrchal – (121006) 1 This printout should have 17 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 2) 10.0 points A 31 m length of coaxial cable has a solid cylindrical wire inner conductor with a di ameter of 2 . 042 mm and carries a charge of 7 . 32 μ C. The surrounding conductor is a cylindrical shell and has an inner diameter of 6 . 748 mm and a charge of 7 . 32 μ C. Assume the region between the conductors is air. The Coulomb constant is 8 . 98755 × 10 9 N · m 2 / C 2 . What is the capacitance of this cable? Correct answer: 1 . 4428 nF. Explanation: Let : k e = 8 . 98755 × 10 9 N · m 2 / C 2 , Q = 7 . 32 μ C , ℓ = 31 m , a = 2 . 042 mm , and b = 6 . 748 mm . The charge per unit length is λ ≡ Q ℓ . V = integraldisplay b a vector E · dvectors = 2 k e λ integraldisplay b a dr r = 2 k e Q ℓ ln parenleftbigg b a parenrightbigg . The capacitance of a cylindrical capacitor is given by C ≡ Q V = ℓ 2 k e 1 ln parenleftbigg b a parenrightbigg = 31 m 2 (8 . 98755 × 10 9 N · m 2 / C 2 ) × 1 ln parenleftbigg 6 . 748 mm 2 . 042 mm parenrightbigg · parenleftbigg 1 × 10 9 nF 1 F parenrightbigg = 1 . 4428 nF 002 (part 2 of 2) 10.0 points What is the potential difference between the two conductors? Correct answer: 5 . 07345 kV. Explanation: Therefore V = Q C = 7 . 32 μ C 1 . 4428 nF parenleftbigg 1 × 10 9 nF 1 F parenrightbigg × parenleftbigg 1 C 1 × 10 6 μ C parenrightbiggparenleftbigg 1 kV 1000 V parenrightbigg = 5 . 07345 kV 003 10.0 points Two conductors insulated from each other are charged by transferring electrons from one conductor to the other. After 9 . 852 × 10 13 have been transferred, the potential difference between the conductors is 11 . 7 V. The charge on an electron is 1 . 60218 × 10 − 19 C . What is the capacitance of the system? Correct answer: 1 . 34911 × 10 − 6 F. Explanation: Let : e = 1 . 60218 × 10 − 19 C , N = 9 . 852 × 10 13 electrons , and V = 11 . 7 V . One conductor gains a charge of Q = N e while the other gains a charge of Q. The capacitance of the system is C = q V = N e V = (9 . 852 × 10 13 ) ( 1 . 60218 × 10 − 19 C) 11 . 7 V = 1 . 34911 × 10 − 6 F . lew (dl9564) – hw 5 – Opyrchal – (121006) 2 004 (part 1 of 2) 10.0 points The potential difference between a pair of oppositely charged parallel plates is 351 V....
View
Full
Document
This note was uploaded on 03/22/2009 for the course PHYS 121 taught by Professor Opyrchal during the Fall '08 term at NJIT.
 Fall '08
 Opyrchal

Click to edit the document details