{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

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

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

View Full Document Right Arrow Icon
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, ( ) ( ) 0 0 0 0 1 1 1 1 , 4 4 4 4 A B q q q q V V y x x y πε πε πε πε = + = + leads to 0 0 0 9 2 2 9 2 2 2 1 1 4 4 4 1 1 2(8.99 10 N m C )(25 10 C) 187 V . 0.80 m 0.60 m B A q q q V V V x y x y πε πε πε 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?
Background image of page 1

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

View Full Document Right Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}