   Chapter 16, Problem 71AP

Chapter
Section
Textbook Problem

Metal sphere A of radius 12.0 cm carries 6.00 μC of charge, and metal sphere B of radius 18.0 cm carries −4.00 μC of charge. If the two spheres are attached by a very long conducting thread, what is the final distribution of charge on the two spheres?

To determine
The final distribution of charge on the spheres.

Explanation

Given info: Sphere A has radius rA=12.0cm and charge qA=6.00μC . Sphere B has radius rB=18.0cm and charge qB=4.00μC , Sphere A has radius rA=12.0cm and charge qA=6.00μC . Sphere B has radius rB=18.0cm and charge qB=4.00μC .

Explanation:

The net charge is,

q=qA+qB

Substitute 6.00μC for qA , 4.00μC for qB

q=(6.00μC)+(4.00μC)=2.00μC

Both the spheres will have the same potential. Therefore,

keqArA=keqBrB

On Re-arranging,

qB=qA(rBrA)

Substitute the above equation the expression for the net charge and re-arrange

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