Question

A large charged conducting sheet is placed in a uniform electric field, perpendicularly to the electric field lines. After placing the sheet into the field, the electric field on the left side of the sheet is $E_1=5\times {10}^5\text{V/m}$ and on the right it is $E_2=3\times {10}^5\text{V/m}$. The sheet experiences a net electric force of $0.08\text{N}$. Find the area of one face of the sheet.$($ Assume the external field to remain constant after introducing the large sheet$)$

$[k=\frac{1}{4\pi {\epsilon }_0}=9\times {10}^9\text{N}{\text{m}}^2{\text{C}}^{-2}]$

• A. $3.6\pi \times {10}^{-2}{\text{m}}^2$
• B. $0.9\pi \times {10}^{-2}{\text{m}}^2$
• C. $1.8\pi \times {10}^{-2}{\text{m}}^2$
• D. None

Answer 1

The correct option is A $3.6\pi \times {10}^{-2}{\text{m}}^2$

Let the external field be $E_0$ and surface charge density of sheet be $\sigma$ (including both surfaces)



Electric field due to induced charge $\sigma$ is given by $$E=\frac{\sigma }{2{\epsilon }_0}$$
So, from the diagram we can deduce that

$E_1=E_0-E$ and $-E_2=E_0+E$

Solving these two equations we get,

$E_0=\frac{E_1-E_2}{2}$ and $E=\frac{-E_1-E_2}{2}$

From the data given in the question,

$E_0=\frac{(5-3)\times {10}^5}{2}={10}^5\text{V/m}$

$E=\frac{(-5-3)\times {10}^5}{2}=-4\times {10}^5\text{V/m}$

and , $\frac{\sigma }{2{\epsilon }_0}=-4\times {10}^5$

$\Rightarrow \sigma =-8{\epsilon }_0\times {10}^5\text{V/m}$

Force on sheet, $$\overrightarrow{F}=q{\overrightarrow{E}}_o$$
$\Rightarrow F=\sigma A{E}_0$

Substituting the given data we get,

$0.08=8{\epsilon }_0\times {10}^5\times A\times {10}^5$

$\Rightarrow A=\frac{{10}^{-12}}{8.85\times {10}^{-12}}\text{or}3.6\pi \times {10}^{-2}{\text{m}}^2$

Hence, option (a) is the correct answer.