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Question
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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Solution
let a be the radius of a sphere A, QA be the charge on the sphere, and CA be the capacitance of the sphere.
Let b be the radius of a sphere B, QB be the charge on the sphere, and CB be the capacitance of the sphere.
Since the two spheres are connected with a wire, their potential (V) will become equal.
Let EA be the electric field of sphere A and EB be the electric field of sphere B.
Therefore, their ratio,
`"E"_"A"/"E"_"B" = "Q"_"A"/(4piin_0 xx "a"_2) xx ("b"^2 xx 4piin_0)/"Q"_"B"`
`"E"_"A"/"E"_"B" = "Q"_"A"/"Q"_"B" xx ("b"^2)/"a"^2` ......(1)
However, `"Q"_"A"/"Q"_"B" = ("C"_"A""V")/("C"_"B""V")`
And, `("C"_"A")/("C"_"B") = "a"/"b"`
∴ `"Q"_"A"/"Q"_"B" = "a"/"b"` .......(2)
Putting the value of (2) in (1), we obtain
∴ `"E"_"A"/"E"_"B" = "a"/"b""b"^2/"a"^2 = "b"/"a"`
Therefore, the ratio of electric fields at the surface is `"b"/"a"`.
A sharp and pointed end can be treated as a sphere of a very small radius and a flat portion behaves as a sphere of a much larger radius. Therefore, the charge density on sharp and pointed ends of the conductor is much higher than on its flatter portions.
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