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Physics Question on Electric charges and fields

A plane square sheet of charge of side 0.5 m0.5\, m has uniform surface charge density. An electron at 1 cm1\, cm from the centre of the sheet experiences a force of 1.6×10−12 N1.6 \times 10^{-12}\, N directed away from the sheet. The total charge on the plane square sheet is (ε0=8.854×10−12C2m−2 N−1)\left(\varepsilon_{0}=8.854 \times 10^{-12} C ^{2} m ^{-2}\, N ^{-1}\right)

A

16.25 μC16.25 \,\mu C

B

−22.15 μC - 22.15 \,\mu C

C

−44.27 μC-44.27 \,\mu C

D

144.27 μC144.27 \,\mu C

Answer

−44.27 μC-44.27 \,\mu C

Explanation

Solution

Given, l=0.5 ml=0.5 \,m

F=1.6×10−12 NF=1.6 \times 10^{-12}\, N
q=−1.6×10−19C   q=-1.6 \times 10^{-19} C \,\,\, (for electron)
ε0=8.854×10−12 C2 m−2 N−1\varepsilon_{0}=8.854 \times 10^{-12} \,C ^{2} \,m ^{-2} \,N ^{-1}
Electric field E=FqE=\frac{F}{q}
or F=qE   ...(i)F=q E \,\,\,...(i)
Electric field due to a square plane sheet of charge
E=σ2ε0E=\frac{\sigma}{2 \varepsilon_{0}}
where σ=\sigma= surface charge density
ε0=\varepsilon_{0}= permittivity of free space
Putting the value of EE in E (i), we get
F=q⋅σ2ε0F =q \cdot \frac{\sigma}{2 \varepsilon_{0}}
σ=2ε0⋅Fq\sigma =2 \varepsilon_{0} \cdot \frac{F}{q}
σ=2×8.854×10−12×1.6×10−12−1.6×10−19\sigma=\frac{2 \times 8.854 \times 10^{-12} \times 1.6 \times 10^{-12}}{-1.6 \times 10^{-19}}
σ=−17.708×10−5\sigma =-17.708 \times 10^{-5}
σA=l2×(−17.708×10−5)\sigma A =l^{2} \times\left(-17.708 \times 10^{-5}\right)
=0.5×0.5×(−17.708×10−5)=0.5 \times 0.5 \times\left(-17.708 \times 10^{-5}\right)
=−4.427×10−5=-4.427 \times 10^{-5}
=−44.27×10−6 C=-44.27 \times 10^{-6}\, C
=−44.27 μ C=-44.27 \,\mu \,C