Question

A plane square sheet of charge of side $0.5\, m$ has uniform surface charge density. An electron at $1\, cm$ 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 $\left(\varepsilon_{0}=8.854 \times 10^{-12} C ^{2} m ^{-2}\, N ^{-1}\right)$

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

Answer 1

The Correct Option is C

Given, $l=0.5 \,m$

$F=1.6 \times 10^{-12}\, N$
$q=-1.6 \times 10^{-19} C \,\,\,$ (for electron)
$\varepsilon_{0}=8.854 \times 10^{-12} \,C ^{2} \,m ^{-2} \,N ^{-1}$
Electric field $E=\frac{F}{q}$
or $F=q E \,\,\,...(i)$
Electric field due to a square plane sheet of charge
$E=\frac{\sigma}{2 \varepsilon_{0}}$
where $\sigma=$ surface charge density
$\varepsilon_{0}=$ permittivity of free space
Putting the value of $E$ in E (i), we get
$ F =q \cdot \frac{\sigma}{2 \varepsilon_{0}} $
$\sigma =2 \varepsilon_{0} \cdot \frac{F}{q}$
$\sigma=\frac{2 \times 8.854 \times 10^{-12} \times 1.6 \times 10^{-12}}{-1.6 \times 10^{-19}}$
$ \sigma =-17.708 \times 10^{-5} $
$ \sigma A =l^{2} \times\left(-17.708 \times 10^{-5}\right) $
$=0.5 \times 0.5 \times\left(-17.708 \times 10^{-5}\right)$
$=-4.427 \times 10^{-5}$
$=-44.27 \times 10^{-6}\, C$
$=-44.27 \,\mu \,C$

Answer 2

We can use Coulomb's law to calculate the force experienced by the electron due to the electric field created by the charged sheet. Coulomb's law states that the force between two charged particles is proportional to the product of their charges and inversely proportional to the square of the distance between them:
\(F = \frac{k \times q_1 \times q_2}{r^2}\)
where \(F\) is the force, \(q_1\) and \(q_2\) are the charges of the two particles, \(r\) is the distance between them, and \(k\) is Coulomb's constant.
In this case, we can consider the electron as a point charge \(q_1\) and the sheet as a continuous distribution of charge with a surface charge density \(σ\). The charge on the sheet is then \(Q = σ \times A\), where \(A\) is the area of the sheet.
The electric field due to the sheet at a distance \(r\) from the center of the sheet is given by:
\(E = \frac{σ}{ (2 \times ε_0)}\)
where \(ε_0\) is the permittivity of free space.
The force experienced by the electron is then:
\(F = q_1 \times E = \frac{q_1 \times σ}{(2 \times ε_0)}\)
Substituting the given values, we have:
\(F = 1.6 \times10^{-12} N\) and \(q_1 = -1.6 \times 10^{-19} C\) (since the force is directed away from the sheet, the electron must have a negative charge) \(r = 0.01 m\) \(σ = \frac{Q}{A}\)
We are asked to find the total charge \(Q\) on the sheet. The area of the sheet is \((0.5 m)^2 = 0.25 m^2.\)
We can rearrange the equation for \(F\) to solve for \(σ\):
\(σ = 2 \times ε_0 \times \frac{F}{q_1}\)
Substituting the given values, we have:
\(σ = 2 \times 8.854 \times \frac{10^{-12} C^2}{m^2N} \times \frac{1.6 \times10^{-12} N}{(-1.6 \times10^{-19} C)} = -1.775 \times10^{-5} C/m^2\)
The total charge on the sheet is then:
\(Q = σ \times A = (-1.775 \times 10^{-5} C/m^2) \times 0.25 m^{2} = -4.4375 \times10^{-6} C\)
The negative sign indicates that the sheet has a net negative charge, which makes sense since the force on the electron is directed away from the sheet.