Question

A large charged metal sheet is placed in a uniform electric field, perpendicular 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 experience a net electric force of 0.08 N. Find the area of one face of the sheet. Assume the external electric field to remain constant after introducing the large sheet. Use $\left(\frac{1}{4{\mathrm{\pi \epsilon }}_0}\right)=9\times {10}^{9}N{m}^2{C}^{-2}$

• 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 of these

Answer 1

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

Let external electric field be $E_0$ and surface charge density of sheet be $\sigma$(Including both surfaces). So,
$E_P=\frac{\sigma }{2{\epsilon }_0}$

Electric field due to sheet
$E_1=E_0-E_P\left(i\right)-E_2=E_0+E_P\left(ii\right)$

From $\left(i\right)$ and $\left(ii\right)$,
$E_0=\frac{E_1-E_2}{2}={10}^{5}V{m}^{-1}\mathrm{and}{E}_P=\frac{-E_1-E_2}{2}=-4\times {10}^{5}V{m}^{-1}\frac{\sigma }{2{\epsilon }_0}=-4\times {10}^{5}or\sigma =-8{\epsilon }_0\times {10}^5\mathrm{Force}\mathrm{on}\mathrm{sheet}F=q{E}_{0}or0.08=\sigma A{E}_{0}or0.08=8{\epsilon }_0\times {10}^{5}A\times {10}^{5}A=\frac{{10}^{-12}}{{\epsilon }_0}={10}^{-12}\times 36\times {10}^9\pi =3.6\pi \times {10}^{-2}{m}^2$