Question

The surface density of charge on the surface of a charged conductor in the air is $\text{26.5}\mu C{m}^{-2}$ . Then the outward force per unit area of the charged conductor is $\left({\epsilon }_0=8.85\times 10^{-12}{\mathrm{C}}^2{\mathrm{N}}^{-1}{\mathrm{m}}^{-2}\right)$

• A. $\text{3.5}N{m}^{-2}$
• B. $\text{17.7}N{m}^{-2}$
• C. $\text{8.85}N{m}^{-2}$
• D. $\text{39.7}N{m}^{-2}$

Answer 1

Answer: (D)

$\sigma =26.5\frac{\mu C}{m^2}=26.5\times 10^{-6}\frac{C}{m^2}$
$\frac{dF}{ds}=\frac{{\sigma }^2}{2k{\epsilon }_0}$
$\frac{dF}{ds}=\frac{26.5\times 26.5\times 10^{-12}}{2\times 1\times \text{8.85}\times 10^{-12}}$ $\dots \left[\because k=1forair\right]$
$=39.67\frac{N}{m^2}\approx 39.7\frac{N}{m^2}$