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Electric force on the charge kept at the centre of a metallic shell
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[QUOTE="Pushoam, post: 6850863, member: 619344"] Applying Gauss's law, $$ \oint (\vec E_{shell} +\vec E_{outside~ charge}) \cdot d\vec A = 0 $$ A conductor has free charge particles which distribute themselves such that 1) net charge and electric field within the meat of the conductor is 0 and the tangential component of electric field on the surface gets 0 as if these being non-zero lead to the movement of electron until they become 0. 2) the component of electric field perpendicular to the surface is nonzero as charged particles can't move in this direction. 3) the conducting surface is equipotential as different potentials at different part of the surface leads to the movement of charge particles such that the potentials become equal. Hence, the electric field at the inner surface of conductor is radially inward as the charges are positive. Since, there is no charged particle within shell, the direction of the electric field inside the shell is unaffected and hence radially inward. Now, $$ \vec A = A \hat r$$ $$\vec E_{shell} +\vec E_{outside~ charge} =( E_{shell} +E_{outside~ charge})(-\hat r)$$ $$ \oint (\vec E_{shell} +\vec E_{outside~ charge}) \cdot d\vec A = 0 $$ $$ \oint -( E_{shell} + E_{outside~ charge}) d A = 0 ~~~ .......(1)$$ If I [B]assume[/B] that ## E_{shell} + E_{outside~ charge} ## is constant over the Gaussian surface, then it has to be 0 for (1) to hold true. [/QUOTE]
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Electric force on the charge kept at the centre of a metallic shell
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