Question

The surface density of a spherical conductor is ${\sigma }_1$, and the electric field on its surface is $E_1$. The surface density of an infinite cylindrical conductor is ${\sigma }_2$ and the electric field on its surface is $E_2$. Which of the following relation is correct ?

• A. $E_1{\sigma }_2=E_2{\sigma }_1$
• B. $E_1{\sigma }_1=E_2{\sigma }_2$
• C. $E_1{\sqrt{\sigma }}_2=E_2{\sqrt{\sigma }}_1$
• D. $E_1{\sqrt{\sigma }}_1=E_2{\sqrt{\sigma }}_2$

Answer 1

Answer: (A)

$E_1=\frac{{\sigma }_1}{{\epsilon }_2}$ also $E_2=\frac{{\sigma }_2}{{\epsilon }_0}$ . Hence $\frac{E_1}{E_2}=\frac{{\sigma }_1}{{\sigma }_2}$