HW-6Solutions-01-23-08

HW-6Solutions-01-23-08 - Physics 21 Spring 2008 HW-6...

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Physics 21 Spring 2008 January 23, 2008 HW-6 Solutions 6-1. (HRW 24-111) Point charges of equal magnitudes (25 nC) and opposite signs are placed on diagonally opposite corners of a rectangle. Point A is the unoccupied corner nearest the positive charge, and point B is the other unoccupied corner. Determine the potential difference V B - V A. Solution: We denote q = 25 × 10 –9 C, y = 0.6 m, x = 0.8 m, with V = the net potential (assuming V 0 as r ). Then, ( ) ( ) 00 11 , 44 AB qq VV y xx y πε =+ leads to 0 92 2 9 222 1 1 4 2(8.99 10 N m C )(25 10 C) 187 V . 0.80 m 0.60 m BA q VV V xy x y ⎛⎞ Δ =−= = ⎜⎟ ⎝⎠ × = 6-2. (HRW 25-1) The two metal objects in Fig. 25-25 have net charges of +70 pC and 70 pC, which result in a 20 V potential difference between them. (a) What is the capacitance of the system? (b) If the charges are changed to +200 pC and 200 pC, what does the capacitance become? (c) What does the potential difference become? F I G U R E 2 5 - 2 5 Solution: (a) The capacitance of the system is C q V == = Δ 70 20 35 pC V pF. .
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This note was uploaded on 02/20/2008 for the course PHYSIC 2 taught by Professor Kim during the Spring '08 term at Lehigh University .

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HW-6Solutions-01-23-08 - Physics 21 Spring 2008 HW-6...

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