extra_credit 02-solutions

extra_credit 02-solutions - howard(cah3459 – extra credit...

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Unformatted text preview: howard (cah3459) – extra credit 02 – turner – (56705) 1 This print-out should have 10 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 Consider two concentric spherical conducting shells. The inner shell has radius a and charge q 1 on it, while the outer shell has radius 3 a and charge q 2 on it. Assume the electric potential V at ∞ is zero. O a 3 a p q 1 q 2 Determine the electric field E at p , where the distance Op = 2 a . 1. E = q 1 20 π a 2 2. E = q 1 4 π a 2 3. E = q 1 8 π a 2 4. E = q 1 18 π a 2 5. E = q 1 16 π a 2 correct 6. E = q 1 6 π a 2 7. E = q 1 6 π a 8. E = q 1 12 π a 2 9. E = q 1 12 π a 10. E = q 1 14 π a 2 Explanation: Set up a Gaussian surface of radius r = 2 a between the shells: r O a 3 a p q 1 q 2 Due to symmetry, the electric field is con- stant over the surface of the sphere, so the flux is simply Φ = E A , and the enclosed charge is q 1 . From Gauss’s Law, Φ = E 4 π r 2 = q 1 4 π E (2 a ) 2 = q 1 E = q 1 4 π (2 a ) 2 = q 1 16 π a 2 . 002 (part 2 of 2) 10.0 points Find the electric potential V at point p . 1. V = 1 4 π q 1 2 a + q 2 3 a correct 2. V = 1 4 π q 1- q 2 a 3. V = 1 4 π q 1 + q 2 a 4. V = 1 4 π q 1 q 2 4 a 2 5. V = 1 4 π q 1 2 a- q 2 3 a 6. V = 1 4 π q 1 + q 2 3 a 7. V = ∞ 8. V = 1 4 π q 1 q 2 a 9. V = 0 10. V = 1 4 π q 1 a + q 2 3 a Explanation: The potential due to a spherical charge dis- tribution is the same as the potential due to howard (cah3459) – extra credit 02 – turner – (56705) 2 a point charge at its center, if V at ∞ is zero (otherwise we would add a constant). Due to q 1 , V 1 = k q 1 r = 1 4 π q 1 2 a . The outer sphere requires more thought. Just outside the surface, the potential due to outer shell is like a point charge at the center; i.e. , at a distance 3 a . As we go inside the shell, Gauss’s Law tells us that since we enclose no charge (we are only considering the outer shell right now), there is no electric field due to outer shell inside the outer shell. If there is no electric field, the potential stays the same as we go inwards. Therefore the potential due to the outer shell at any point inside the outer shell is V 2 = k q 2 3 a = 1 4 π q 2 3 a and V = V 1 + V 2 = 1 4 π q 1 2 a + q 2 3 a ....
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This note was uploaded on 11/28/2010 for the course PHY 56705 taught by Professor Turner during the Spring '10 term at University of Texas.

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extra_credit 02-solutions - howard(cah3459 – extra credit...

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