Question

Two large conducting sheet are placed parallel to each other with a separation of $2\text{cm}$ between them. An electron starting from rest near one of the sheets reaches the other plate in $2\mu \text{s}$. The surface charge density on the inner surface is
[Assume both sheets have same charge density]

• A. $3.1\times {10}^{-13}{\text{C/m}}^2$
• B. $4.1\times {10}^{-13}{\text{C/m}}^2$
• C. $5.1\times {10}^{-13}{\text{C/m}}^2$
• D. $6.1\times {10}^{-13}{\text{C/m}}^2$

Answer 1

The correct option is C $5.1\times {10}^{-13}{\text{C/m}}^2$

Given:
$s=2\text{cm};t=2\times {10}^{-6}\text{s}$


The two plates should have opposite nature of charge distribution else net electric field inside the two sheets will be zero.

Now, net electric field inside the sheet is given by

$E=\frac{\sigma }{2{ϵ}_0}+\frac{\sigma }{2{ϵ}_0}$

$\Rightarrow E=\frac{\sigma }{{ϵ}_0}$ (towards right)

Further,
force on electron due to electric field,

$F=qE=q\frac{\sigma }{{ϵ}_0}$ (towards left)

So, acceleration,

$a=\frac{F}{m}=\frac{q\sigma }{{ϵ}_{0}m}$

From equation of motion,

$s=ut+\frac{1}{2}a{t}^2$

Since, electron start from rest. So, $u=0$

$\Rightarrow s=\frac{1}{2}\times \frac{q\sigma }{{ϵ}_{0}m}\times t^2$

$\Rightarrow \sigma =\frac{2{ϵ}_{0}ms}{q{t}^2}$

$\Rightarrow \sigma =\frac{2\times 9\times {10}^{-12}\times 9.1\times {10}^{-31}\times 2\times {10}^{-2}}{1.6\times {10}^{-19}\times (2\times {10}^{-6}{)}^2}$

$\Rightarrow \sigma =5.1\times {10}^{-13}{\text{C/m}}^2$

So, option $\left(c\right)$ is correct.