But since areas differ by a factor of 4, the effective ratio simplifies to 2 : 1.
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प्रश्न
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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उत्तर
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
संबंधित प्रश्न
A spherical conducting shell of inner radius r1 and outer radius r2 has a charge Q.
(a) A charge q is placed at the centre of the shell. What is the surface charge density on the inner and outer surfaces of the shell?
(b) Is the electric field inside a cavity (with no charge) zero, even if the shell is not spherical, but has any irregular shape? Explain.
(a) Show that the normal component of electrostatic field has a discontinuity from one side of a charged surface to another given by
`(vec"E"_2 - vec"E"_1).hat"n" = sigma/in_0`
Where `hat"n"` is a unit vector normal to the surface at a point and σ is the surface charge density at that point. (The direction of `hat"n"` is from side 1 to side 2.) Hence show that just outside a conductor, the electric field is σ `hat"n"/in_0`
(b) Show that the tangential component of electrostatic field is continuous from one side of a charged surface to another.
[Hint: For (a), use Gauss’s law. For, (b) use the fact that work done by electrostatic field on a closed loop is zero.]
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