extra_credit 02-solutions

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

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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 – fnd 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 p q 1 q 2 Determine the electric feld E at p , where the distance Op =2 a . 1. E = q 1 20 π± 0 a 2 2. E = q 1 4 0 a 2 3. E = q 1 8 0 a 2 4. E = q 1 18 0 a 2 5. E = q 1 16 0 a 2 correct 6. E = q 1 6 0 a 2 7. E = q 1 6 0 a 8. E = q 1 12 0 a 2 9. E = q 1 12 0 a 10. E = q 1 14 0 a 2 Explanation: Set up a Gaussian surFace oF radius r a between the shells: r O p q 1 q 2 Due to symmetry, the electric feld is con- stant over the surFace oF the sphere, so the ±ux is simply Φ = EA , and the enclosed charge is q 1 . ²rom Gauss’s Law, Φ = E 4 π r 2 = q 1 ± 0 4 π E (2 a ) 2 = q 1 ± 0 E = q 1 4 0 (2 a ) 2 = q 1 16 0 a 2 . 002 (part 2 oF 2) 10.0 points ²ind the electric potential V at point p . 1. V = 1 4 0 ± q 1 2 a + q 2 3 a ² correct 2. V = 1 4 0 ³ q 1 - q 2 a ´ 3. V = 1 4 0 ³ q 1 + q 2 a ´ 4. V = 1 4 0 ± q 1 q 2 4 a 2 ² 5. V = 1 4 0 ± q 1 2 a - q 2 3 a ² 6. V = 1 4 0 ³ q 1 + q 2 3 a ´ 7. V = 8. V = 1 4 0 ± q 1 q 2 a ² 9. V =0 10. V = 1 4 0 ± q 1 a + q 2 3 a ² Explanation: The potential due to a spherical charge dis- tribution is the same as the potential due to

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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 = kq 1 r = 1 4 π± 0 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 Feld due to outer shell inside the outer shell. If there is no electric Feld, 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 = 2 3 a = 1 4 0 q 2 3 a and V = V 1 + V 2 = 1 4 0 ± q 1 2 a + q 2 3 a ² . 003 10.0 points A charge of 2 . 8 μ C is 20 cm above the center of a square of side length 40 cm. The permittivity of free space is 8 . 85 × 10 - 12 C 2 / N · m 2 .
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extra_credit 02-solutions - howard(cah3459 extra credit 02...

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