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# exam1sol - Flores Michael – Exam 1 – Due 11:00 pm –...

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Unformatted text preview: Flores, Michael – Exam 1 – Due: Feb 17 2004, 11:00 pm – Inst: Sonia Paban 1 This print-out should have 16 questions. Multiple-choice questions may continue on the next column or page – find all choices before making your selection. The due time is Central time. 001 (part 1 of 2) 10 points Assume: The electric potential V = 0 at ∞ . Consider two concentric spherical conduct- ing 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. 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 ² a 2 2. E = q 1 16 π ² a 2 correct 3. E = q 1 6 π ² a 2 4. E = q 1 14 π ² a 2 5. E = q 1 8 π ² a 2 6. E = q 1 12 π ² a 2 7. E = q 1 4 π ² a 2 8. E = q 1 18 π ² 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 ² 4 π E (2 a ) 2 = q 1 ² E = q 1 4 π ² (2 a ) 2 = q 1 16 π ² a 2 . 002 (part 2 of 2) 10 points Find the electric potential V at point p . 1. V = 1 4 π ² ‡ q 1 a + q 2 3 a · 2. V = 1 4 π ² µ q 1- q 2 a ¶ 3. V = 1 4 π ² ‡ q 1 2 a- q 2 3 a · 4. V = 1 4 π ² µ q 1 + q 2 3 a ¶ 5. V = 1 4 π ² ‡ q 1 2 a + q 2 3 a · correct 6. V = 1 4 π ² µ q 1 + q 2 a ¶ 7. V = 0 8. V = ∞ 9. V = 1 4 π ² ‡ q 1 a + q 2 4 a · 10. V = 1 4 π ² ‡ 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 Flores, Michael – Exam 1 – Due: Feb 17 2004, 11:00 pm – Inst: Sonia Paban 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 ² 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 ² q 2 3 a and V = V 1 + V 2 = 1 4 π ² ‡ q 1 2 a + q 2 3 a · . 003 (part 1 of 1) 10 points A system of charges are shown in the figure. ‘ 2 ‘ 2 ‘ + Q + Q- Q What is the electric potential energy of the following system of charges. 1. U = k Q 2 2 ‘ 2. U = k Q 2 ‘ 2 3. U = k Q 2 3 ‘ 4. U =- k Q 2 ‘ 5. U =- k Q 2 3 ‘ 6. None of these....
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## This note was uploaded on 02/18/2010 for the course PHYS 303l taught by Professor Panab during the Spring '04 term at North Texas.

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exam1sol - Flores Michael – Exam 1 – Due 11:00 pm –...

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