Solution 9 - homework 09 KANUNGO ANIL Due 4:00 am Question...

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homework 09 – KANUNGO, ANIL – Due: Sep 25 2007, 4:00 am 1 Question 1, chap 26, sect 1. part 1 of 1 10 points Two large parallel conducting plates P and Q are connected to a battery of emf E , as shown. A test charge is placed successively at points I , II , and III . E + - I II III If edge effects are negligible, the force on the charge when it is at point III is 1. of equal magnitude and in the same di- rection as the force on the charge when it is at point I . 2. equal in magnitude to the force on the charge when it is at point I , but in the oppo- site direction. 3. much greater in magnitude than the force on the charge when it is at point II , but in the same direction. 4. much less in magnitude than the force on the charge when it is at point II , but in the same direction. 5. of equal magnitude and in the same di- rection as the force on the charge when it is at point II . correct Explanation: Neglecting edge effects, the electric field strength between the two plates is uniform. Therefore the force on the test charge is the same at points II and III , both in magnitude and in direction. When edge effects are negligible, the elec- tric field strength at point I is zero, so the force on the test charge is zero when it is at point I . Question 2, chap 26, sect 1. part 1 of 1 10 points A parallel plate capacitor is connected to a battery. + Q - Q d 2 d If we double the plate separation, 1. the charge on each plate is halved. cor- rect 2. the electric field is doubled. 3. the potential difference is halved. 4. the capacitance is doubled. 5. None of these. Explanation: The capacitance of a parallel plate capaci- tor is C = ǫ 0 A d . Hence doubling d halves the capacitance, and Q = C V is also halved parenleftbigg C = ǫ 0 A 2 d = 1 2 ǫ 0 A d = 1 2 C parenrightbigg . Question 3, chap 26, sect 1.
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