This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: guajardo (jmg4256) Ch17h3extra chiu (57425) 1 This printout should have 17 questions. Multiplechoice 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 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. 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) Ch17h3extra 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
 Spring '08
 Turner

Click to edit the document details