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Unformatted text preview: sarceno (dea457) – Ch17h3extra – yao – (57465) 1 This printout should have 17 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Consider a parallel plate capacitor system with plate charge Q ( Q > 0) and cross sec tion A of each plate. Top plate Bottom plate + Q − Q Plate area is A d E gap Denote the magnitude of the attractive force by which the charge of the top plate pulls on the charge of the bottom plate by F , and denote the electric field vector due to the charges on the top plate by vector E top . Find i) the magnitude F of the force; ii) the magnitude of vector E top at the bottom plate; and iii) the direction of vector E top . ↓ indicates downward and ↑ indicates upward. 1. F = Q 2 2 ǫ A , E top = Q ǫ A , vector E top bardbl vector E top bardbl ↑ 2. F = Q 2 ǫ A , E top = Q 2 ǫ A , vector E top bardbl vector E top bardbl ↓ 3. F = Q 2 ǫ A , E top = Q ǫ A , vector E top bardbl vector E top bardbl ↓ 4. F = Q 2 2 ǫ A , E top = Q ǫ A , vector E top bardbl vector E top bardbl ↓ 5. F = Q 2 2 ǫ A , E top = Q 2 ǫ A , vector E top bardbl vector E top bardbl ↓ correct 6. F = Q 2 2 ǫ A , E top = Q 2 ǫ A , vector E top bardbl vector E top bardbl ↑ 7. F = Q 2 ǫ A , E top = Q ǫ A , vector E top bardbl vector E top bardbl ↑ 8. F = Q 2 ǫ A , E top = Q 2 ǫ A , vector E top bardbl vector E top bardbl ↑ Explanation: Recall from chapter 16 that the magni tude of the electric field between two oppo sitely charged parallel conducting plates of area A with charge magnitude  Q  is given by  vector E cap  = Q/ǫ A , and that this is a su perposition of the (equal) contributions of both plates. Therefore, the magnitude of the field due to the top plate must be given by  vector E top  = Q/ 2 ǫ A . By definition, this field must point from the positive to the negative plate, so the direction is downward . Finally, the force is simply obtained from F = qE , so F = Q 2 / 2 ǫ A . 002 10.0 points Given two parallel plate capacitors 1 and 2. They both have a platecharge Q and plate area A. The gap of 1 is filled with a dielectric medium with dielectric constant K 1 = 2 and the gap of 2 is filled with a dielectric constant K 2 = 4. Determine the ratio of the polarized charge q 1 q 2 . 1. 4 9 2. 4 3 3. 1 3 4. 1 5. 8 9 6. 7 9 7. 2 9 8. 5 9 9. 2 3 correct 10. 1 9 Explanation: sarceno (dea457) – Ch17h3extra – yao – (57465) 2 K = E E ′ = Q Q − q q = Q parenleftbigg 1 − 1 K parenrightbigg For capacitor 1, we know that K 1 = 2 q 1 = Q parenleftbigg 1 − 1 2 parenrightbigg = Q 2 For capacitor 2, we know that K 2 = 4 q 2 = Q parenleftbigg 1 − 1 4 parenrightbigg = 3 Q 4 q 1 q 2 = ( Q/ 2) (3 Q/ 4) = 2 3 003 10.0 points Consider a conducting sphere with radius R and charge + Q, surrounded by a conducting spherical shell with inner radius 2 R , outer radius 3 R and net charge + Q....
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This note was uploaded on 11/12/2012 for the course PHY 303L taught by Professor Turner during the Spring '08 term at University of Texas.
 Spring '08
 Turner
 Physics

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