Ch21-p007 - 7 We assume the spheres are far apart Then the charge distribution on each of them is spherically symmetric and Coulomb’s law can be

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Unformatted text preview: 7. We assume the spheres are far apart. Then the charge distribution on each of them is spherically symmetric and Coulomb’s law can be used. Let q 1 and q 2 be the original charges. We choose the coordinate system so the force on q 2 is positive if it is repelled by q 1 . Then, the force on q 2 is F q q r k q q r a = − = − 1 4 1 2 2 1 2 2 π ε where r = 0.500 m. The negative sign indicates that the spheres attract each other. After the wire is connected, the spheres, being identical, acquire the same charge. Since charge is conserved, the total charge is the same as it was originally. This means the charge on each sphere is ( q 1 + q 2 )/2. The force is now one of repulsion and is given by F r k q q r b q q q q = = + + + 1 4 4 2 2 2 1 2 2 2 1 2 1 2 π ε d i d i b g . We solve the two force equations simultaneously for q 1 and q 2 . The first gives the product q q r F k a 1 2 2 2 9 12 0500 0108 8 99 10 300 10 = − = − × ⋅ = − × − . ....
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This note was uploaded on 05/21/2010 for the course PHYS 231 taught by Professor Kd during the Spring '10 term at The Petroleum Institute.

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Ch21-p007 - 7 We assume the spheres are far apart Then the charge distribution on each of them is spherically symmetric and Coulomb’s law can be

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