In a region, the intensity of an electric field is given by $E = 2 i + 3 j + k$ in $N C^{- 1}$ . The electric flux through a surface $S = 10 i m^{2}$ in the region is:

5 N m^{2} C^{- 1}

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10 N m^{2} C^{- 1}

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15 N m^{2} C^{- 1}

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20 N m^{2} C^{- 1}

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Solution

The correct option is C $20 N m^{2} C^{- 1}$
Given: $E = (2 i + 3 j + k) N C^{- 1}$ and $S = 10 i m^{2}$
We know that:
$ϕ = E \cdot S$
$ϕ = (2 i + 3 j + k) \cdot (10 i)$
$ϕ = 20 N m^{2} C^{- 1}$

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In a region, the intensity of an electric field is given by E=2i+3j+k in NC−1. The electric flux through a surface S=10i m2 in the region is:

The correct option is C 20 Nm2C−1Given: E=(2i+3j+k)NC−1 and S=10i m2We know that:ϕ=E⋅Sϕ=(2i+3j+k)⋅(10i)ϕ=20 Nm2C−1

In a region, the intensity of an electric field is given by $E = 2 i + 3 j + k$ in $N C^{- 1}$ . The electric flux through a surface $S = 10 i m^{2}$ in the region is:

The correct option is C $20 N m^{2} C^{- 1}$
Given: $E = (2 i + 3 j + k) N C^{- 1}$ and $S = 10 i m^{2}$
We know that:
$ϕ = E \cdot S$
$ϕ = (2 i + 3 j + k) \cdot (10 i)$
$ϕ = 20 N m^{2} C^{- 1}$