Q.The electric field intensity outside the charged conducting sphere of radius 'R', placed in a medium of permittivity $\in$ at a distance 'r' from the centre of the sphere in terms of surface charge density $\sigma$ is

A uniformly charged spherical shell behaves, for points outside it, exactly like a point charge at its centre; the field inside the shell is zero.

For a thin spherical shell of radius $R$ carrying total charge $q$, uniformly distributed over its surface, Gauss's law gives three regions:

• Outside the shell ($r > R$): $\displaystyle E = \frac{1}{4\pi\varepsilon_0}\frac{q}{r^2} = \frac{kq}{r^2}$, directed radially (as if all charge were concentrated at the centre).
• On the surface ($r = R$): $\displaystyle E = \frac{q}{4\pi\varepsilon_0 R^2}$. …

For a charged conducting sphere of surface charge density $\sigma$, the field outside follows from Gauss's law, $E=\dfrac{Q}{4\pi\epsilon r^2}$ with $Q=\sigma\times4\pi R^2$.

For a conducting sphere of radius $R$ with uniform surface charge density $\sigma$, total charge $Q = \sigma\times4\pi R^2$. By Gauss's law, the field at a distance $r\,(>R)$ from the centre, in a medium of permittivity $\in$, is the same as that of a point charge $Q$ at the centre: …

$E=\dfrac{Q}{4\pi\varepsilon_0\varepsilon_r r^2}$; check which listed change actually reduces $E$.

The electric intensity at a point outside a uniformly charged sphere, at distance $r$ from its centre, in a medium of relative permittivity (dielectric constant) $\varepsilon_r$, is

$E=\frac{Q}{4\pi\varepsilon_0\varepsilon_r r^2}$

Checking each option for whether it decreases $E$:

• (A) Increase in charge $Q$: $E\propto Q$, so this increases $E$ — not the answer.
• (B) Increase in dielectric constant $\varepsilon_r$: since $E\propto 1/\varepsilon_r$, increasing $\varepsilon_r$ decreases $E$ — matches.
• (C) Decrease in distance $r$ from the centre: $E\propto 1/r^2$, so decreasing $r$ increases $E$ — not the answer. …