Question

A hollow metal sphere of radius $R$ is uniformly charged. The electric field due to the sphere at , distance r from the centre :

• A. zero as r increases for r < R. increases as. r increases for r > R
• B. decreases as r increases for r < R and r > R
• C. increases as r increases for r < R and for r > R
• D. zero as r increases for r < R. decreases r as increases for r > R

Answer 1

The Correct Option is D

To solve this problem, we need to consider the behavior of the electric field due to a uniformly charged hollow metal sphere. The electric field behavior can be explained using Gauss's Law, which relates the electric field to the charge enclosed by a Gaussian surface.

1. First, consider the case when the distance from the center of the sphere, $r$, is less than the radius of the sphere, $R$ (i.e., $r < R$):
   • Inside a hollow charged sphere, the electric field is zero at any point. This is because the net charge enclosed by a Gaussian surface inside the hollow sphere is zero.
   • Therefore, the electric field $E$ remains zero as $r$ increases within the hollow sphere.
2. Next, consider the case when $r$ is greater than the radius of the sphere, $R$ (i.e., $r > R$):
   • Outside a uniformly charged sphere, the electric field behaves as if all the charge were concentrated at the center of the sphere. The magnitude of the electric field at a distance $r$ from the center is given by Coulomb's Law:
   • E = \frac{1}{4\pi\epsilon_0} \frac{Q}{r^2}, where $Q$ is the total charge and $\epsilon_0$ is the permittivity of free space.
   • This formula shows that as $r$ increases, $E$ decreases, since $E$ is inversely proportional to $r^2$.

Based on this understanding, the correct explanation of the behavior of the electric field is: "zero as $r$ increases for $r < R$. decreases as $r$ increases for $r > R$."