# HW4c - GenericStudent Homework 4 savrasov 39823 This...

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

GenericStudent – Homework 4 – savrasov – 39823 – Jan 02, 2008 1 This print-out should have 11 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 1) 1 points A 18 pF capacitor is connected across a 68 V source. What charge is stored on it? Correct answer: 1 . 224 × 10 - 9 C. Explanation: Let : C = 18 pF = 1 . 8 × 10 - 11 F and V = 68 V . The capacitance is C = q V q = C V = (1 . 8 × 10 - 11 F) (68 V) = 1 . 224 × 10 - 9 C keywords: 002 (part 1 of 2) 1 points An isolated conducting sphere can be consid- ered as one element of a capacitor (the other element being a concentric sphere of infinite radius). If k = 1 4 π ² 0 , the capacitance of the system is C , and the charge on the sphere is Q , what is the radius r of the sphere? 1. r = C k 2. r = k Q C 3. r = k C correct 4. r = Q 2 C k 5. r = k Q 6. r = k C Q Explanation: The definition of capacitance is C = Q V Q = C V . Let V ( ) = 0. Then the potential of the conducting sphere is V = k Q r = k C V r , so r = k C . 003 (part 2 of 2) 1 points The Coulomb constant is 8 . 98755 × 10 9 N · m 2 / C 2 . If the potential on the surface of the sphere is 5700 V and the capacitance is 7 . 56 × 10 - 11 F, what is the surface charge density? Correct answer: 74 . 278 nC / m 2 . Explanation: Let : k = 8 . 98755 × 10 9 N · m 2 / C 2 , V = 5700 V , and C = 7 . 56 × 10 - 11 F . The charge is Q = C V . Assuming a uniform surface charge density we have σ = Q 4 π r 2 = C V 4 π ( k C ) 2 = 1 4 π k 2 V C = 1 4 π (8 . 98755 × 10 9 N · m 2 / C 2 ) 2 × 5700 V 7 . 56 × 10 - 11 F = 74 . 278 nC / m 2 .

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

View Full Document
GenericStudent – Homework 4 – savrasov – 39823 – Jan 02, 2008 2 keywords: 004 (part 1 of 2) 1 points A capacitor network is shown below. 85 V 20 μ F 24 μ F 4 μ F a b Find the equivalent capacitance C ab be- tween points a and b for the group of ca- pacitors. Correct answer: 14 . 9091 μ F. Explanation: Let : C 1 = 20 μ F , C 2 = 24 μ F , C 3 = 4 μ F , and E B = 85 V . E B C 1 C 2 C 3 a b For capacitors in series, 1 C series = X 1 C i V series = X V i , and the individual charges are the same. For parallel capacitors, C parallel = X C i Q parallel = X Q i , and the individual voltages are the same.
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern