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(D) If Coulomb's force varies as $F \propto 1/r^3$,then the electric field $E$ at a distance $r$ from a point charge $q$ is given by $E = k q / r^3$.F

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field inside will be zero and charge density inside will not be zero

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(D) If Coulomb's force varies as $F \propto 1/r^3$,then the electric field $E$ at a distance $r$ from a point charge $q$ is given by $E = k q / r^3$.For a solid metallic sphere of radius $R$ with total charge $Q$,by symmetry,the electric field inside the sphere $(r However,Gauss's law is modified for this force law. The flux $\phi$ through a spherical Gaussian surface of radius $r$ $(r In the standard case $(F \propto 1/r^2)$,Gauss's law states $\oint E \cdot dA = q_{enclosed} / \epsilon_0$. If the force law changes to $1/r^3$,the relationship between flux and enclosed charge changes. Specifically,for a $1/r^3$ force,the flux through a closed surface is not simply proportional to the enclosed charge. Calculations show that for this force law,the charge density $\rho$ inside the conductor is non-zero to maintain $E=0$ inside,as the standard Gauss's law does not hold in the same way. Thus,the field inside remains zero due to symmetry,but the charge density inside is non-zero.

If some charge is given to a solid metallic sphere,the field inside remains zero and by Gauss's law all the charge resides on the surface. Now,suppose that Coulomb's force between two charges varies as $1 / r^{3}$. Then,for a charged solid metallic sphere

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