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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 surface of 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 fatter portions ?

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Answer Text

Solution :
Let a be the radius of a sphere A, `Q_(A)` be the charge on the sphere, and `C_(A)` be the capacitance of the sphere. Let b be the radius of a sphere B, `Q_(B)` be the charge on the sphere, and `C_(B)` be the capacitance of the sphere. Since the two spheres are connected with a wire, their potential (V) will become equal. <br> Let `E_(A)` be the electric field of sphere A and `E_(B)` be the electric field of sphere B. Therefore, their ratio, <br> `(E_(A))/(E_(B)) = (Q_(A))/(4pi in_(0) xx a_(2)) xx (b^(2) xx 4pi in_(0))/(Q_(B))` <br> `(E_(A))/(E_(B)) = (Q_(A))/(Q_(B)) xx (b^(2))/(a^(2))` ... (i) <br> However, `(Q_(A))/(Q_(B)) = (C_(A)V)/(C_(B)V)` <br> And, `(C_(A))/(C_(B)) = (a)/(b)` <br> `:. (Q_(A))/(Q_(B)) = (a)/(b)` ...(2) <br> Putting the value of (2) in (1), we obtain <br> `:. (E_(A))/(E_(B)) -(a)/(B)(b^(2))/(a^(2)) = (b)/(a)` <br> Therefore, the ratio of electric fields at the surface is `(b)/(a)`.

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