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प्रश्न
Fill in the blank.
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_________.
उत्तर
Let the point charge be q.
by gauss's law, the charge on the inner surface will be - q
Surface charge density of the inner surface `sigma_i = - q/(4pir^2)`
by charge conservation, on the hollow sphere, the outer surface will have charge q
Surface charge density of the inner surface `sigma_o = q/(4pi(2r)^2) = q/(16pir^2)`
ratio = `sigma_i/sigma_o = (-q/(4pir^2))/(q/(16pir^2)) = -4/1`
APPEARS IN
संबंधित प्रश्न
(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.]
If Coulomb’s law involved 1/r3 dependence (instead of 1/r2), would Gauss’s law be still true?
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