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Let's say that there's a hollow conducting sphere placed in the presence of an external electric field and there's a + ve charge placed inside the sphere at a point other than the centre of the sphere, now my question is whether the charge will experience '0 net electic force ' or not and why? also tell me about the forces on it due to induced charges.

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The field that a charge creates inside a conducting sphere can be obtained by the method of images. The result is that there is will be a force acting on the charge, whose magnitude is proportional to $$F\propto \frac{q^2 R d}{\left(R^2-d^2\right)^2}\ ,$$ where $R$ is the internal radius of the sphere and $d$ is the distance of the charge from the sphere's center. The proportionality constant depends on if you work in cgs or MKS. The full derivation can be found here.

Note that this force results from the interaction of the charge with the sphere, and is not due to the external field, as @PrishChakraborty correctly states, because of the Faraday Cage effect (however, he is wrong in saying that the "the charge will not experience any force").

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The presence of the charge inside should indeed induce a negative charge on the interior surface of the sphere. However, this will lead to the generation of a constant electrostatic potential through the interior of the surface. Since the action of a force depends on a difference of potential through \begin{equation}F=-q\frac{dV}{dr} \end{equation} the charge will not experience any force. As for the external electric field, while it will affect the conducting sphere, due the aforementioned principle it will not have a direct influence on the charge inside. You may read further about a Faraday Cage.

I think, however, the electric field might produce a force on the conducting sphere and result in exerting a contact force on the charge inside but I'm not sure so I won't talk about it.

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  • $\begingroup$ You are right about the Farady Cage, but wrong about the vanishing of the force. I added an answer that explains this. $\endgroup$ – yohBS Jul 2 '15 at 10:12
  • $\begingroup$ @yohBS Ah right, sorry, it was silly of me to not think about the force due to the induced charge. Thanks for clearing it up! $\endgroup$ – Prish Chakraborty Jul 2 '15 at 18:54

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