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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

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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

But since areas differ by a factor of 4, the effective ratio simplifies to 2 : 1.

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2019-2020 (March) Delhi Set 2

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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(Take `1/(4 pi epsilon_0` = 9 × 109 SI units)


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