Question

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

Answer 1

Answer: (B)

increase in dielectric constant

The electric field at a point outside a sphere of charge is given by

$E=\frac{1}{4\pi {ϵ}_0\kappa }\frac{Q}{r^2}$

where $Q$ is the charge on the sphere,

$\kappa$ is the dielectric constant,

$r$ is the distance from the centre of the sphere.

From the expression for electric field, we get

$E\propto Q$

Thus, as the charge on sphere increases, the intensity of the electric field increases.

Similarly, $E\propto \frac{1}{\kappa }$

Thus, as the dielectric constant increases, the intensity of the electric field decreases.

Similarly, $E\propto \frac{1}{r^2}$

Thus, as $r$ or $r^2$ decreases, the intensity of electric field increases.