Solution 9 - homework 09 – KANUNGO, ANIL – Due: Sep 25...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 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 b b 2 d b b 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 A d . Hence doubling d halves the capacitance, and Q = C V is also halved parenleftbigg C ′ = ǫ A 2 d = 1 2 ǫ A d = 1 2 C parenrightbigg . Question 3, chap 26, sect 1....
View Full Document

This homework help was uploaded on 04/02/2008 for the course PHY 303L taught by Professor Turner during the Fall '08 term at University of Texas.

Page1 / 5

Solution 9 - homework 09 – KANUNGO, ANIL – Due: Sep 25...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online