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Question
State and explain Gauss’s law.
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Solution
Gauss’s law states that the flux of the electric field through any closed surface S is 1/∈ₒ times the total charge enclosed by S
Let the total flux through a sphere of radius r enclose a point charge q at its centre. Divide the sphere into a small area element as shown in the figure.

The flux through an area element ΔS is
`Deltaphi=E.DeltaS=q/(4piin_0r^2)hatr.DeltaS`
Here, we have used Coulomb’s law for the electric field due to a single charge q.
The unit vector `hatr`is along the radius vector from the centre to the area element. Because the normal to a sphere at every point is along the radius vector at that point, the area element ΔS and `hatr` have the same direction. Therefore
`Deltaphi=q/(4piin_0r^2)DeltaS`
Because the magnitude of the unit vector is 1, the total flux through the sphere is obtained by adding the flux through all the different area elements.
`phi=sum_(all DeltaS)q/(4piin_0r^2)DeltaS`
Because each area element of the sphere is at the same distance r from the charge,
`phi=q/(4piin_0r^2)sum_(all DeltaS)DeltaS=q/(4piin_0r^2)S`
Now, S the total area of the sphere equals 4πr². Thus,
`pi=q/(4piin_0r^2)xx4pir^2=q/in_0`
Hence, the above equation is a simple illustration of a general result of electrostatics called Gauss’s law
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Consider a region inside which there are various types of charges but the total charge is zero. At points outside the region
- the electric field is necessarily zero.
- the electric field is due to the dipole moment of the charge distribution only.
- the dominant electric field is `∞ 1/r^3`, for large r, where r is the distance from a origin in this region.
- the work done to move a charged particle along a closed path, away from the region, will be zero.




