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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. Suppose now that coulomb’s force between two charges varies as $\dfrac{1}{{{r^3}}}$. Then, for a charged solid metallic sphere
(A) field inside will be zero and charge density inside will be zero
(B) field inside will not be zero and charge density inside will not be zero
(C) field inside will not be zero and charge density inside will be zero
(D) field inside will be zero and charge density will not be zero

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Last updated date: 26th Feb 2024
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Hint: When the charge is given to the metallic sphere, the electric field inside the sphere will be definitely zero in all conditions. But the charge density is varying depending on the amount of charge given to the solid metallic sphere.

Complete step by step solution
Given that,
The coulomb’s force between the two charges varies as $\dfrac{1}{{{r^3}}}$.
By Gauss’s law,
If the Coulomb's force is directly proportional to the $\dfrac{1}{{{r^3}}}$ is not valid.
The coulomb force is inversely proportional to the square of the distance, but here it is given that the coulomb force is inversely proportional to the cube of the distance. So, it is not valid.
$\phi \ne \dfrac{Q}{{{\varepsilon _0}}}$
This is a condition for the given Coulomb's force value.
Where, $\phi $ is the electric flux, $Q$ is the charge enclosed and ${\varepsilon _0}$ is the permittivity of free space.
For the electrostatic conditions the electric field is zero for both of conductor, $E = 0$
But, the charge density inside the metallic sphere is not equal to zero. It has some other values, when the charge is given to the solid metallic sphere.
Thus, the field inside the metallic sphere will be zero and the charge density of the metallic sphere will not be zero.

Hence, the option (D) is the correct answer.

Note: The electric flux of the system is directly proportional to the charge supplied to the system and the electric flux is inversely proportional to the permittivity of the free space. As the charge increases, the electric flux also increases, as the permittivity of the free space increases, the electric flux decreases.