Q.Electric intensity due to a charged sphere at a point outside the sphere decreases with ______ (A) increase in charge on sphere
(B) increase in dielectric constant
(C) decrease in the distance from the centre of sphere
(D) Decrease in square of distance from the centre of sphere

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. …