But since areas differ by a factor of 4, the effective ratio simplifies to 2 : 1.
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Question
A point charge is placed at the centre of a hollow conducting sphere of internal radius ‘r’ and outer radius ‘2r’. The ratio of the surface charge density of the inner surface to that of the outer surface will be ______.
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Solution
A point charge is placed at the centre of a hollow conducting sphere of internal radius ‘r’ and outer radius ‘2r’. The ratio of the surface charge density of the inner surface to that of the outer surface will be 2 : 1.
Explanation:
When a point charge is placed at the centre of a hollow conducting sphere, an equal and opposite charge is induced on the inner and outer surfaces.
Since surface charge density (σ) = `Q/(4 pi R^2)`, it is inversely proportional to the square of radius.
Inner radius = r
Outer radius = 2r
`sigma_"inner" : sigma_"outer" = 1/r^2 : 1/((2 r)^2)`
= 4 : 1
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