Consider a uniform electric field $E = 3 \times 10^{3} \hat{i} N / C .$ (a) What is the flux of this field through a square of 10 cm on a side whose plane is parallel to the yz plane? (b) What is the flux through the same square if the normal to its plane makes a 60° angle with the x-axis?

Solution

Given, Electric field, \vec{E} = 3 \times 10^{3} \hat{i} {NC}^{- 1}

(a)

Area of the square = 10 \times 10 = 100 {cm}^{2} = 10^{- 2} m^{- 2}

The area of a surface can be represented as a vector along normal to the surface. Since normal to the square is along x-axis, we have

∆ \vec{S} = 10^{- 2} \hat{i} m^{- 2}

Electricity flux through the square

= (3 \times 10^{3} \hat{i}) . (10^{- 2} \overset{⏜}{i}) = 30 {Nm}^{2} C^{- 1}

(b) Given, the angle between area vector and the electric field is 60°.

Therefore,

Electric flux - ϕ = \vec{E} . ∆ \vec{S} = E . ds \cos 60 ° = 3 \times 10^{3} \times 10^{- 2} \times \frac{1}{2} = 15 {Nm}^{2} C^{- 1}

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