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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
field inside will be zero and charge density inside will be zero
field inside will not be zero and charge density inside will not be zero
field inside will not be zero and charge density inside will be zero
field inside will be zero and charge density inside will not be zero
Solution
$(d)$ As Coulomb's force,
$\quad F \propto \frac{1}{r^{3}}$
$\Rightarrow \text { Electric field, } E \propto \frac{1}{r^{3}} \Rightarrow E=\frac{k q}{r^{3}}$
Now, for Gaussian surface $(r < R)$.
Electric flux linked with surface is
$\phi=\int \frac{k q}{r^{3}} \cdot 2 \pi r d r$
$=2 \pi k q \int \frac{d r}{r^{2}}=2 \pi k q\left(-\frac{1}{r}\right)$
As flux is non-zero, charge density is also non-zero. Also, by symmetry of charge distribution electric field is zero.
