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Question: A circular plane sheet of radius 10 cm is placed in a uniform electric field of \(5 \times 10^{5} ⥂ ...

A circular plane sheet of radius 10 cm is placed in a uniform electric field of 5×105⥂NC−15 \times 10^{5} ⥂ NC^{- 1}, making an angle of 60o60^{o}with the field. The electric flux through the sheet is.

A

1.36×102Nm2C−11.36 \times 10^{2}Nm^{2}C^{- 1}

B

1.36×104Nm2C−11.36 \times 10^{4}Nm^{2}C^{- 1}

C

0.515×102Nm2C−10.515 \times 10^{2}Nm^{2}C^{- 1}

D

0.515×104Nm2C−10.515 \times 10^{4}Nm^{2}C^{- 1}

Answer

1.36×104Nm2C−11.36 \times 10^{4}Nm^{2}C^{- 1}

Explanation

Solution

: Here, r = 10 cm = 0.1 m

E=5×105NC−1E = 5 \times 10^{5}NC^{- 1}

As the angle between the plane sheet and the electric field is 60o,60^{o}, angle made by the normal to the plane sheet and the electric field is

θ=90o−60o=30o\theta = 90^{o} - 60^{o} = 30^{o}

φE=EScos⁡θ=E×πr2cos⁡θ\varphi_{E} = ES\cos\theta = E \times \pi r^{2}\cos\theta

=5×105×3.14×(0.1)2cos⁡30o= 5 \times 10^{5} \times 3.14 \times (0.1)^{2}\cos 30^{o}

=1.36×104Nm2C−1= 1.36 \times 10^{4}Nm^{2}C^{- 1}