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InterviewSolution
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Two charged conducting spheres of radill a and b are connected to eachother by a wire. What is the ratio of electric fields at the surface of two spheres ? Use the result obtained to explain why charge density on the sharp and pointed ends of a conducter is higher than on its fatter portions ? |
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Answer» The charge flows from the sphere at higher potential to the other at lower potential, till their potentials become equal. After sharing, the charges on two spheres would be `(Q_(1))/(Q_(2)) = (C_(1) V)/(C_(2) V)` where `C_(1), C_(2)` are the capaciteres of two spheres. But `(C_(1))/(C_(2)) = (a)/(b) :. (Q_(1))/(Q_(2)) = (a)/(b)` Ratio of surface density of charge on the two spheres: `(sigma_(1))/(sigma_(2)) = (Q_(1))/(4pi a^(2)) . (4pi b^(2))/(Q^(2)) = (Q_(1))/(Q_(2)) . (b^(2))/(a^(2)) = (a)/(b) (b^(2))/(a^(2)) = (b)/(a)` Hence ratio of electric fields at the surfaces of two spheres `(E_(1))/(E_(2)) = (sigma_(1))/(sigma_(2)) = (b)/(a)` A sharp and pointed end can be treated as a sphere of very small radius and a flat portion behaves as a sphere of much larger radius. Therefore, charge density on sharp and pointed ends of conductor is much higher than on its flatter portions. |
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