Question

A large charged metal 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 $E_1=5\times {10}^{5}V{m}^{-1}$ and on the right it is $E_2=3\times {10}^{5}V{m}^{-1}$ . The sheet experiences a net electric force of 0.08 N. Find the area of one face of the sheet. Assume the external field to remain constant after introducing the large sheet :

Use $\left(\frac{1}{4\pi {\epsilon }_0}\right)=9\times {10}^{9}N{m}^2{C}^{-2}$

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

Answer 1

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

$F=q{E}_0$ or $0.08=\sigma A{E}_0$Let external field be $E_0$ and surface charge density of sheet be $\sigma$ (including both surface). So
$E_p=\frac{\sigma }{2{\epsilon }_0}$
Electric field due to sheet
$E_1=E_0-E_p$ (i)
$-E_2=E_0+E_p$(ii)
From (i) and (ii),
$E_0=\frac{E_1-E_2}{2}={10}^{5}V{m}^{-1}$
$E_p=\frac{-E_1-E_2}{2}=-4\times {10}^{5}V{m}^{-1}$
$\frac{\sigma }{2{\epsilon }_0}=-4\times {10}^5$ or $\sigma =-8{\epsilon }_0\times {10}^5$
Force on sheet
$F=q{E}_0$ or $0.08=\sigma A{E}_0$
or $A=\frac{{10}^{-12}}{{\epsilon }_0}={10}^{-12}\times 36\times {10}^9\pi =3.6\pi \times {10}^{-2}{m}^2$