Question

A plane square sheet of charge of side $0.5m$ has uniform surface charge density. An electron at $1cm$ from the centre of the sheet experiences a force of $1.6\times 10^{-12}N$ directed away from the sheet. The total charge on the plane square sheet is $\left({\epsilon }_0=8.854\times 10^{-12}{C}^2{m}^{-2}{N}^{-1}\right)$

• A. $16.25\mu C$
• B. $-22.15\mu C$
• C. $-44.27\mu C$
• D. $144.27\mu C$
• E. $8.854\mu C$

Answer 1

Answer: (C)

Given, $l=0.5m$

$F=1.6\times 10^{-12}N$
$q=-1.6\times 10^{-19}C$ (for electron)
${\epsilon }_0=8.854\times 10^{-12}{C}^2{m}^{-2}{N}^{-1}$
Electric field $E=\frac{F}{q}$
or $F=qE...(i)$
Electric field due to a square plane sheet of charge
$E=\frac{\sigma }{2{\epsilon }_0}$
where $\sigma =$ surface charge density
${\epsilon }_0=$ permittivity of free space
Putting the value of $E$ in Eq. (i), we get
$F=q\cdot \frac{\sigma }{2{\epsilon }_0}$
$\sigma =2{\epsilon }_0\cdot \frac{F}{q}$
$\sigma =\frac{2\times 8.854\times 10^{-12}\times 1.6\times 10^{-12}}{-1.6\times 10^{-19}}$
$\sigma =-17.708\times 10^{-5}$
$\sigma A=l^2\times \left(-17.708\times 10^{-5}\right)$
$=0.5\times 0.5\times \left(-17.708\times 10^{-5}\right)$
$=-4.427\times 10^{-5}$
$=-44.27\times 10^{-6}C$
$=-44.27\mu C$