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Unformatted text preview: gonzales (pag757) – Homework 5 – Hoffmann – (56745) 1 This print-out should have 11 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A 67 m length of coaxial cable has a solid cylindrical wire inner conductor with a di- ameter of 1 . 311 mm and carries a charge of 10 . 88 μ C. The surrounding conductor is a cylindrical shell and has an inner diameter of 11 . 287 mm and a charge of- 10 . 88 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 . 73136 nF. Explanation: Let : k e = 8 . 98755 × 10 9 N · m 2 / C 2 , Q = 10 . 88 μ C , ℓ = 67 m , a = 1 . 311 mm , and b = 11 . 287 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 = 67 m 2 (8 . 98755 × 10 9 N · m 2 / C 2 ) × 1 ln parenleftbigg 11 . 287 mm 1 . 311 mm parenrightbigg · parenleftbigg 1 × 10 9 nF 1 F parenrightbigg = 1 . 73136 nF 002 10.0 points A parallel-plate capacitor is charged by con- necting it to a battery. If the battery is disconnected and the sep- aration between the plates is increased, what will happen to the charge on the capacitor and the electric potential across it? 1. The charge remains fixed and the electric potential increases. correct 2. The charge increases and the electric po- tential remains fixed. 3. The charge and the electric potential de- crease. 4. The charge decreases and the electric po- tential increases. 5. The charge increases and the electric po- tential decreases. 6. The charge and the electric potential in- crease. 7. The charge and the electric potential re- main fixed. 8. The charge remains fixed and the electric potential decreases. 9. The charge decreases and the electric po- tential remains fixed. Explanation: Charge is conserved, so it must remain con- stant since it is stuck on the plates. With the battery disconnected, Q is fixed. C = ǫ A d A larger d makes the fraction smaller, so C is smaller. Thus the new potential V ′ = Q C ′ is larger. gonzales (pag757) – Homework 5 – Hoffmann – (56745) 2 003 10.0 points A capacitor is constructed of interlocking plates as shown in the figure (a cross-sectional view). The separation between adjacent plates is 0 . 49 mm, and the effective area of one of the adjacent plates is 10 . 8 cm 2 ....
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This note was uploaded on 01/27/2012 for the course PHY 303L taught by Professor Turner during the Spring '08 term at University of Texas.
- Spring '08