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Unformatted text preview: oldhomewk 10 LEE, BENJAMIN Due: Sep 26 2007, 4:00 am 1 Question 1, chap 26, sect 1. part 1 of 1 10 points A small electrically charged object is sus pended by a thread between the vertical plates of a parallelplate capacitor. The acceleration of gravity is 9 . 8 m / s 2 . A B vector E + E 530 mg 12 nC =13 6 cm What is the potential difference between the plates? Correct answer: 5 . 99565 kV (tolerance 1 %). Explanation: Let : m = 530 mg = 0 . 00053 kg , g = 9 . 8 m / s 2 , d = 6 cm = 0 . 06 m , = 13 , and q = 12 nC = 1 . 2 10 8 C . A B vector E + E m q d The forces acting on the particle are the tension T on the thread, the weight mg and the electric force q E = q V d . Since the par ticle is in equilibrium, the resultant force is equal to zero; i.e. , T sin = q V d T cos = mg Dividing, tan = q V mg d V = mg d tan q = (0 . 00053 kg) (9 . 8 m / s 2 ) (0 . 06 m) 1 . 2 10 8 C tan 13 = 5 . 99565 kV . Question 2, chap 26, sect 2. part 1 of 2 10 points A 9 . 22 F capacitor is first charged by being connected across a 10 V battery. It is then disconnected from the battery and connected across an uncharged 9 . 11 F capacitor. Find the charge on the 9 . 22 F capacitor. Correct answer: 46 . 3767 C (tolerance 1 %). Explanation: oldhomewk 10 LEE, BENJAMIN Due: Sep 26 2007, 4:00 am 2 Let : C 1 = 9 . 22 F , C 2 = 9 . 11 F and V = 10 V . The initial charge of each capacitor is Q 1 = C 1 V , and Q 2 = 0 . After the capacitors are connected in paral lel, the potential difference across each is the same, so V = Q 1 C 1 = Q 2 C 2 Q 2 = Q 1 C 2 C 1 ....
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This note was uploaded on 04/07/2008 for the course PHY 303K taught by Professor Turner during the Fall '07 term at A.T. Still University.
 Fall '07
 Turner
 Charge, Acceleration

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