PhyIIHW4Solutions - hopkins (tlh982) – HW04 – criss –...

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Unformatted text preview: hopkins (tlh982) – HW04 – criss – (4908) 1 This print-out should have 24 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 A coaxial cable has a charged inner conductor (with charge +5 . 3 μ C and radius 1 . 415 mm) and a surrounding oppositely charged con- ductor (with charge- 5 . 3 μ C and radius 7 . 412 mm). The Coulomb constant is 8 . 98755 × 10 9 N · m 2 / C 2 . Assume the region between the conductors is air, and neglect end effects. 2 8 m 1 . 415 mm +5 . 3 μ C 7 . 412 mm- 5 . 3 μ C b What is the magnitude of the electric field halfway between the two cylindrical conduc- tors? Correct answer: 7 . 70914 × 10 5 V / m. Explanation: Let : k e = 8 . 98755 × 10 9 N · m 2 / C 2 , a = 1 . 415 mm = 0 . 001415 m , b = 7 . 412 mm = 0 . 007412 m , Q = 5 . 3 μ C = 5 . 3 × 10 − 6 C , and ℓ = 28 m . ℓ r = a + Q r = b- Q b The linear charge density is λ = Q ℓ = 5 . 3 × 10 − 6 C 28 m = 1 . 89286 × 10 − 7 C / m . The halfway point is r = a + b 2 = . 001415 m + 0 . 007412 m 2 = 0 . 0044135 m . The electric field of a cylindrical capacitor is given by E = 2 k e λ r = 2 ( 8 . 98755 × 10 9 N · m 2 / C 2 ) . 0044135 m × (1 . 89286 × 10 − 7 C / m) = 7 . 70914 × 10 5 V / m . 002 (part 2 of 2) 10.0 points What is the capacitance of this cable? Correct answer: 0 . 940663 nF. Explanation: The charge per unit length is λ ≡ Q ℓ . Since 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 hopkins (tlh982) – HW04 – criss – (4908) 2 = 28 m 2 (8 . 98755 × 10 9 N · m 2 / C 2 ) × 1 ln parenleftbigg 7 . 412 mm 1 . 415 mm parenrightbigg parenleftbigg 1 × 10 9 nF 1 F parenrightbigg = . 940663 nF . 003 10.0 points A 1-megabit computer memory chip contains many 37 fF capacitors. Each capacitor has a plate area of 3 . 24 × 10 − 11 m 2 . The characteristic atomic diameter is 10 − 10 m = 1 ˚ A. The permittivity of a vac- uum is 8 . 85419 × 10 − 12 C 2 / N · m 2 . Determine the plate separation of such a capacitor (assume a parallel-plate configura- tion). Correct answer: 77 . 534 ˚ A. Explanation: Let : A = 3 . 24 × 10 − 11 m 2 , C = 37 fF = 3 . 7 × 10 − 14 F , and ǫ = 8 . 85419 × 10 − 12 C 2 / N · m 2 . d = ǫ A C = ( 8 . 85419 × 10 − 12 C 2 / N · m 2 ) 3 . 7 × 10 − 14 F × ( 3 . 24 × 10 − 11 m 2 ) = 7 . 7534 × 10 − 9 m = 77 . 534 ˚ A ....
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PhyIIHW4Solutions - hopkins (tlh982) – HW04 – criss –...

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