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Question

Physics Question on Electric charges and fields

What is the electric flux linked with closed surface ?

A

1011Nm2C−110^{11}N m^{2}C^{-1}

B

1012Nm2C−110^{12}N m^{2}C^{-1}

C

1010Nm2C−110^{10}N m^{2}C^{-1}

D

8.86×1013Nm2C−18.86\times10^{13} N m^{2 }C^{-1}

Answer

1012Nm2C−110^{12}N m^{2}C^{-1}

Explanation

Solution

According to Gauss?? law Electric flux, ϕ=qε0\phi=\frac{q}{\varepsilon_{0}} where, q = total charge enclosed by closed surface ∴ ϕ=1.25+7+1−0.4ε0=8.85C8.85×10−12C2N−1m−2=1012Nm2C−1\therefore\, \phi=\frac{1.25+7+1-0.4}{\varepsilon_{0}} =\frac{8.85 C}{8.85\times10 ^{-12} C^{2}N^{-1}m^{-2}}=10^{12}N m^{2} C^{-1}