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Question
Two charged conducting spheres of radii a and b are connected to each other by a wire. Find the ratio of the electric fields at their surfaces.
Solution
Let a be the radius of a sphere A, QA be the charge on the sphere and CA be the capacitance of that sphere. Let b be the radius of a sphere B, QB be the charge on the sphere, and CB be the capacitance of that sphere. Since the two spheres are connected by a wire, their potential (V) will become equal. Let EA be the electric field of sphere A and EB by the electric field of sphere B. Then,
`E_A/E_B = ((Q_A)/(4piε_0a^2)) xx ((b^2 4piε_0)/(Q_B))`
`E_A/E_B = Q_A/Q_B xx b^2/a^2` ......................(i)
However, `Q_A/Q_B = (C_AV)/(C_BV)` .............(ii)
and `C_A/C_B = a/b` ...........(iii)
Putting the values of (ii) and (iii) in (i), we get,
`E_A/E_B = b/a`
Therefore, the required ratio of electric fields at the surface of the spheres is `b/a`.
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