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Question: A large plate sheet of charge having surface charge density \(50 \times 10^{- 16}Cm^{- 2}\)lies in t...

A large plate sheet of charge having surface charge density 50×10−16Cm−250 \times 10^{- 16}Cm^{- 2}lies in the x-y plane. The electric flux through a circular area of radius 0.10.1 m is, if the normal to the circular area makes an angle of 60o60^{o}with the z – axis.

A

4.4×10−6Nm2C−14.4 \times 10^{- 6}Nm^{2}C^{- 1}

B

2.2×10−7Nm2C−12.2 \times 10^{- 7}Nm^{2}C^{- 1}

C

4.4×10−7Nm2C−14.4 \times 10^{- 7}Nm^{2}C^{- 1}

D

2.2×10−7Nm2C−12.2 \times 10^{- 7}Nm^{2}C^{- 1}

Answer

4.4×10−7Nm2C−14.4 \times 10^{- 7}Nm^{2}C^{- 1}

Explanation

Solution

: Here, σ=5.0×10−16Cm−2\sigma = 5.0 \times 10^{- 16}Cm^{- 2}

r1=0.1m,θ=60or_{1} = 0.1m,\theta = 60^{o}

Field due to a plane sheet of charge, , E=σ2ε0E = \frac{\sigma}{2\varepsilon_{0}}

Flux through circular area,

φE=EΔScos⁡θ=σ2ε0×πr2cos⁡θ\varphi_{E} = E\Delta S\cos\theta = \frac{\sigma}{2\varepsilon_{0}} \times \pi r^{2}\cos\theta

=5.0×10−16×3.14×(0.1)2cos⁡60o2×8.85×10−12= \frac{5.0 \times 10^{- 16} \times 3.14 \times (0.1)^{2}\cos 60^{o}}{2 \times 8.85 \times 10^{- 12}}

=4.4×10−7⥂Nm2C−1= 4.4 \times 10^{- 7} ⥂ Nm^{2}C^{- 1}