The electric field inside a conductor must be zero. Can this be explained with the help of Gauss's law?

Also, how does this explain that if we have a conducting sphere with a cavity inside, and we place a point charge (say positive, at the centre of this cavity) then there is a negative charged induced on the inner surface of this sphere, and a positive charge induced on the outer surface of the sphere?


1 Answer 1


Yes it can be explained using gauss's law, if you assume that charge on conductors reside on its surface.

You could take a random gaussian surface inside the conductor and use gauss's law i.e. $$\oint \vec E \cdot d \vec A = \cfrac{q_{inside}}{\epsilon_0}$$ Since all the charge on the conductor resides on surface of the conductor $q_{inside} = 0$. $$\therefore \oint \vec E \cdot d \vec A = \cfrac{0}{\epsilon_0} = 0$$ Since area of conductor isn't zero $\vec E = 0$.



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