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 $\left(\varepsilon_0=8.85 \times 10^{-12} \quad \mathrm{C}^2 \quad \mathrm{~N}^{-1} \quad \mathrm{~m}^{-2}\right)$

• A. $\text{3.5}Nm^{- 2}$
• B. $\text{17.7}Nm^{- 2}$
• C. $\text{8.85}Nm^{- 2}$
• 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}} \, $
$\frac{dF}{ds}= \, \frac{\sigma ^{2}}{2 kε_{0}}$
$\frac{dF}{ds}=\frac{26.5 \times 26.5 \, \times \, 10^{-12}}{2 \, \times \, 1 \, \times \, \text{8.85} \, \times \, 10^{- 12}} \, $ $\ldots \left[\because k = 1 \, for \, air\right]$
$ \, \, \, =39.67 \, \frac{N}{m^{2}} \, \approx 39.7 \, \frac{N}{m^{2}}$