मराठी

A Thin Conducting Spherical Shell of Radius R Has Charge Q Spread Uniformly Over Its Surface. Using Gauss’S Law, Derive an Expression for an Electric Field at a Point Outside the Shell.

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प्रश्न

A thin conducting spherical shell of radius R has charge Q spread uniformly over its surface. Using Gauss’s law, derive an expression for an electric field at a point outside the shell.

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उत्तर

According to Gauss law,

`epsi_0EointdA=q`

Where,

q is the point charge

E is electric field due to the point charge

dA is a small area on the Gaussian surface at any distance and

`epsi_0` is the proportionality constant

For a spherical shell at distance r from the point charge, the integral `ointdA` is merely the sum of all differential of dA on the sphere.

Therefore, `oint dA =4pir^2`

`epsi_0E(4pir^2) = q`

or `,E = q/(epsi4pir^2)`

Therefore, for a thin conducting spherical shell of radius R and charge Q, spread uniformly over its surface, the electric field at any point outside the shell is

`E = Q/(e_0 4pir^2)`

Where r is the distance of the point from the centre of the shell.

`E = q/(4piepsi_0r^2)`

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2008-2009 (March) Delhi set 1

संबंधित प्रश्‍न

State Gauss’s law for magnetism. Explain its significance. 


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The surface considered for Gauss’s law is called ______.


Five charges q1, q2, q3, q4, and q5 are fixed at their positions as shown in figure. S is a Gaussian surface. The Gauss’s law is given by `oint_s E.ds = q/ε_0`

Which of the following statements is correct?


If `oint_s` E.dS = 0 over a surface, then ______.

  1. the electric field inside the surface and on it is zero.
  2. the electric field inside the surface is necessarily uniform.
  3. the number of flux lines entering the surface must be equal to the number of flux lines leaving it.
  4. all charges must necessarily be outside the surface.

If there were only one type of charge in the universe, then ______.

  1. `oint_s` E.dS ≠ 0 on any surface.
  2. `oint_s` E.dS = 0 if the charge is outside the surface.
  3. `oint_s` E.dS could not be defined.
  4. `oint_s` E.dS = `q/ε_0` if charges of magnitude q were inside the surface.

Consider a region inside which there are various types of charges but the total charge is zero. At points outside the region

  1. the electric field is necessarily zero.
  2. the electric field is due to the dipole moment of the charge distribution only.
  3. the dominant electric field is `∞ 1/r^3`, for large r, where r is the distance from a origin in this region.
  4. the work done to move a charged particle along a closed path, away from the region, will be zero.

Refer to the arrangement of charges in figure and a Gaussian surface of radius R with Q at the centre. Then

  1. total flux through the surface of the sphere is `(-Q)/ε_0`.
  2. field on the surface of the sphere is `(-Q)/(4 piε_0 R^2)`.
  3. flux through the surface of sphere due to 5Q is zero.
  4. field on the surface of sphere due to –2Q is same everywhere.

In the diagram, the total electric flux through the closed surface ‘S’ is:

[Given q = charge ε0 = permittivity of free space]


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