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 Question Redistribution of Charges ( Physics Forums Homework Help Archive )Updated: 2009-01-01 07:40:30 (5)
 Redistribution of Charges Identical isolated conducting spheres 1 and 2 have equal amounts of charge and are separated by a distance large compared with their diameters (Fig. 22-20a). The electrostatic force acting on sphere 2 due to sphere 1 is F. Suppose now that a third identical sphere 3, having an insulating handle and initially neutral, is touched first to sphere 1 (Fig. 22-20b), then to sphere 2 (Fig. 22-20c), and finally removed (Fig. 22-20d). In terms of F, what is the electrostatic force F' that now acts on sphere 2? The way I did it, when Sphere1 touches Sphere3, they both end up w/ electrostatic force of 1/2F. Then when Sphere3 touches Sphere2, they distribute evenly to have a electrostatic force of 0.75 F. But this answer is wrong, what did i do wrong?
 Answers: Redistribution of Charges ( Physics Forums Homework Help Archive )
Redistribution of Charges

 Originally Posted by CinderBlockFist OK, first i had particle1 had -1 charge, and neutral particle 3 had +0.5, and -0.5 charge (since it is the same size as particle 1). When they touch, I added up the charges and divided by two, since the charges would distribute equally, so both end up w/ -0.75 and +0.25 charge.
Just worry about the - charge. + charge is a deficit or lack of - charge.

Sphere 1 has charge -1. S3 has 0 charge. When s3 touches it, charges flow so that the electrical potentials are equal. Since they are identical spheres, this means that each sphere has the same charge. So s3 and s1 each have -1/2 charge. The same thing happens when s3 touches s2. The net charge (+1/2) is distributed equally between s2 and s3 so each charge ends up with +.25.

So you end up with s1 having -.5 charge and s2 having a + .25. Since the coulomb force is proportional to the product of the charges, the new force will be: F' = .125 F

AM

Andrew Mason

 Redistribution of Charges OH crap, thanks mason, leme try it out. CinderBlockFist
 Redistribution of Charges OK, first i had particle1 had -1 charge, and neutral particle 3 had +0.5, and -0.5 charge (since it is the same size as particle 1). When they touch, I added up the charges and divided by two, since the charges would distribute equally, so both end up w/ -0.75 and +0.25 charge. Next, particle 3 touches 2, so i get net of -0.75, and +1.25, dividing by 2, I get +0.625 and -0.375, for a total net charge of +0.25 for particles 2 and 3. Why is this answer wrong? My final 2 answers available are 0.5 and 0.375 CinderBlockFist
 Redistribution of Charges Use the expression of Coulomb force and take into account that,due to equal potential,the spheres that come into contact have the same charge... Daniel. dextercioby
Redistribution of Charges

 Originally Posted by CinderBlockFist The way I did it, when Sphere1 touches Sphere3, they both end up w/ electrostatic force of 1/2F. Then when Sphere3 touches Sphere2, they distribute evenly to have a electrostatic force of 0.75 F.
Be careful about the sign of the charges. A +1/2 F charge on sphere 3 and -1F charge on 2 gives a net -1/2 F shared between 2 and 3.

AM

Andrew Mason

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