Question

8. State Coulomb’s law and express it in vector form. Derive it using Gauss theorem.

Answer 1

Draw a Gaussian sphere of radius r at the centre of which charge +q is located

All the points on this surface are equivalent. Due to symmetry, the electric field E has the same magnitude at every point on the surface of the sphere and it is radially outward in direction. Therefore, for a area element dS around any point P on the Gaussian surface both E and dS are directed radially outward, Hence, the angle between E and dS is zero.

The flux passing through the area element dS ,that is,

d φ =E.dS= EdS cos 00=EdS

Hence, the total flux through the entire Gaussian sphere is obtained as,

Φ=∫EdS =E∫dS

But ∫dS is the total surface area of the sphere and is equal to 4πr2,

Φ=E(4πr2)                                                      ------Eqn (1)

But according to Gauss law,

Φ=q/ε0                                                         ------Eqn (2) where q is the charge enclosed within the closed surface

By comparing equation (1) and (2) ,we get E(4πr2)=q/ε0

Hence, E=q/4πε0r2 -----------------Eqn(3)

The equation (3) is the expression for the magnitude of the intensity of electric field E at a point which is at distant r from the point charge +q.

In a second point charge q0 be placed at the point at which the magnitude of E is computed, then the magnitude of the force acting on the second charge q0 would be

F=q0E

By substituting value of E from equation (3), we get F=qoq/4πε0r2 ------------Eqn(4)

The equation (4) represents the Coulomb’s Law and it is derived from gauss law.