A spherical charged conductor has surface density of charge as $\sigma $. The electric field intensity on its surface is $E$. If the radius of the surface is doubled, keeping $\sigma $ unchanged, what will be the electric field intensity on the new sphere?
$\text{A}\text{. }\dfrac{E}{2}$
$\text{B}\text{. }\dfrac{E}{4}$
$\text{C}\text{. }2E$
$\text{D}\text{. }E$

Hint: We will calculate the initial and final value of electric field intensity of the charged sphere. The radius of the sphere is doubled, keeping the surface charge density unchanged so we will find the relation how electric field depends upon the radius of the sphere and its surface charge density.

Formula used:
$E=\dfrac{\sigma }{{{\varepsilon }_{o}}}$

Complete step by step answer:
Electric field intensity at a point is described as the force experienced by a unit positive charge placed at that point. Electric field intensity is a vector quantity, meaning it has both magnitude and the direction.
Expression for Electric field intensity,
$E=\dfrac{F}{q}$
Where,
$E$ is the electric field intensity
$F$ is the force experienced by the charged particle
$q$ is the amount of charge carried by the particle
 For a spherical charged conductor, electric field intensity on its surface is given as,
$E=\dfrac{\sigma }{{{\varepsilon }_{o}}}$
Where,
$E$ is the electric field density of sphere
$\sigma $ is the surface charge density of sphere
${{\varepsilon }_{o}}$ is the permittivity of free space
It is given that the radius of sphere has been doubled, keeping the surface charge density unchanged,
Electric field intensity of a charged sphere does not depend on the radius of the sphere,
Therefore, electric field intensity $E$ will remain unchanged.
Hence, the correct option is D.

Note: The electric field intensity of a charged sphere does not depend upon the radius of the sphere but depends upon the surface charge density of the sphere. The charge on a sphere is considered to be distributed uniformly on its surface which gives a constant value of surface charge density. Even if we change the radius of the sphere, be it smaller or bigger, the surface charge density of the sphere remains unchanged.