**preview**has

**blurred**sections. Sign up to view the full version! View Full Document

**Unformatted text preview: **guajardo (jmg4256) - Ch17-h3-extra - chiu - (57425) 1 This print-out should have 17 questions. Multiple-choice questions may continue on the next column or page - find all choices before answering. 001 0.0 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 A , E top = Q A , vector E top bardbl vector E top bardbl 2. F = Q 2 A , E top = Q A , vector E top bardbl vector E top bardbl 3. F = Q 2 2 A , E top = Q 2 A , vector E top bardbl vector E top bardbl correct 4. F = Q 2 A , E top = Q A , vector E top bardbl vector E top bardbl 5. F = Q 2 A , E top = Q 2 A , vector E top bardbl vector E top bardbl 6. F = Q 2 2 A , E top = Q 2 A , vector E top bardbl vector E top bardbl 7. F = Q 2 2 A , E top = Q A , vector E top bardbl vector E top bardbl 8. F = Q 2 A , E top = Q 2 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/ 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 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 A . 002 0.0 points Given two parallel plate capacitors 1 and 2. They both have a plate-charge 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. 1 2. 5 9 3. 8 9 4. 2 9 5. 4 9 6. 1 9 7. 4 3 8. 2 3 correct 9. 1 3 10. 7 9 Explanation: guajardo (jmg4256) - Ch17-h3-extra - chiu - (57425) 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 0.0 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....

View Full Document