Question

In the following figure an isolated charged conductor is shown. The correct statement will be: -

• A. $E_{\text{A}}>E_{\text{B}}>E_{\text{C}}>E_{\text{D}}$
• B. $E_{\text{A}}<E_{\text{B}}<E_{\text{C}}<E_{\text{D}}$
• C. $E_{\text{A}}=E_{\text{B}}=E_{\text{C}}=E_{\text{D}}$
• D. $E_{\text{B}}=E_{\text{C}}\text{and}{E}_{\text{A}}>E_{\text{D}}$

Answer 1

The correct option is A $E_{\text{A}}>E_{\text{B}}>E_{\text{C}}>E_{\text{D}}$

As we know that the electric field intensity for conductor is given by,

$E=\frac{\sigma }{2{\epsilon }_0}$

Intensity of electric field on sharp edges is more, because$\sigma \propto \frac{1}{r}$

As for sharp edge $r$ is very less
therefore $\sigma$ is more hence $E$ is more.

$\therefore E_{\text{A}}>E_{\text{B}}>E_{\text{C}}>E_{\text{D}}$