Drive coulomb inverse square law with the help of gauss theorem?
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Answer:
Coulomb Inverse Square Law
To derive Coulomb’s Law from gauss law or to find the intensity of electric field due to a point charge +q at any point in space using Gauss’s law ,draw a Gaussian sphere of radius r at the centre of which charge +q is located.
For a area element dS around any point P on the Gaussian surface both E and dS are directed radially outward,that is ,the angle between E and dS is zero
The flux passing through the area element dS ,that is,
d φ =E.dS= EdS cos 0°= EdS
Therefore flux through entire Gaussian sphere is
Φ=∫EdS
But ∫dS is the total surface area of the sphere and is equal to 4πr2,that is,
Φ=E(4πr²)
But according to Gauss’s law for electrostatics
Φ=q/ε0
Where q is the charge enclosed within the closed surface
By comparing above equations ,we get
E(4πr²) =q/ε0
As q0 is in electric field,
F=qE substituting this gives
F=qo q/4πε0 r²
Hope it helps!
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