Question

In the given figure, ${\sigma }_1$ and ${\sigma }_2$ are the surface charge densities around points $1$ and $2$ respectively. Then the relation between the electric fields $E_1$ and $E_2$ near the points $1$ and $2$ respectively will be :

• A. $E_1>E_2$
• B. $E_1<E_2$
• C. $E_1=E_2$
• D. can't be interpreted

Answer 1

The correct option is B $E_1<E_2$

We know that surface charge density $\sigma$, of a conductor at a particular region on its surface is inversely proportional to the radius of curvature $r_c$.
i.e. $\sigma \propto \frac{1}{r_c}$


In the given problem, since $(r_c{)}_1>(r_c{)}_2$
So, we can conclude that,
${\sigma }_2>{\sigma }_1$

We know that electric field near a charged conductor surface is given by :
$E=\frac{\sigma }{{ϵ}_o}$

$\because {\sigma }_2>{\sigma }_1$

$\therefore E_2>E_1$