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Two charged conducting spheres of radii a and b are connected to each other by a wire. What is the ratio of electric fields at the surfaces of the two spheres? Use the result obtained to explain why charge density on the sharp and pointed ends of a conductor is higher than on its flatter portions. |
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Answer» Solution :With arbitrary q and Q FRO A and B respectively. `E_(A)/E_(B)=((1)/(4pi epsi_(0)) q/a^(2))/((1)/(4pi epsi_(0)) Q/b^(2))=q/Q . b^(2)/a^(2)` For equilaterial CONDITION, `V_(A)=V_(B)` `q/(4pi epsi_(0) a)=(Q)/(4pi epsi_(0)b) therefore q/Q=q/b ""E_(A)/E_(B)=a/b. b^(2)/a^(2)=b/a` `E=sigma/(epsi_(0))`. For pointed ends `sigma=q/A`. As `A to 0` E becomes EXTREMELY large. Flat MEANS `A to oo` hence. 
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