The electric field in a region is given by E -35 E 0 i -45 E 0 j - with E 0-2 - 0 - 103 N C -1- Find the flux of this field through a rectangular surface of area 0-2 m 2 parallel to the y-z plane-

Given: Electric field strength, E #8594;=35E0i^+45 #160;E0j #9180;, where E0 #160;= #160;2.0 #215;103 #160;N/C The plane of the rectangular surfa ...

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Byju's Answer

Standard XII

Physics

Idea of Symmetry

The electric ...

Question

The electric field in a region is given by $\vec{E} = \frac{3}{5} E_{0} \vec{i} + \frac{4}{5} E_{0} \vec{j} with E_{0} = 2 \cdot 0 \times 10^{3} N C^{- 1}$ . Find the flux of this field through a rectangular surface of area 0⋅2 m² parallel to the y-z plane.

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Solution

Given:
Electric field strength, $\vec{E} = \frac{3}{5} E_{0} \hat{i} + \frac{4}{5} E_{0} \overset{⏜}{j}$ ,
where $E_{0} = 2.0 \times 10^{3} N / C$

The plane of the rectangular surface is parallel to the y-z plane. The normal to the plane of the rectangular surface is along the x axis. Only $\frac{3}{5} E_{0} \overset{⏜}{i}$ passes perpendicular to the plane; so, only this component of the field will contribute to flux.

On the other hand, $\frac{4}{5} E_{0} \overset{⏜}{j}$ moves parallel to the surface.
Surface area of the rectangular surface, a = 0⋅2 m²

Flux,

$ϕ = \vec{E} . \vec{a} = E \times a ϕ = (\frac{3}{5} \times 2 \times 10^{3}) \times (2 \times 10^{- 1}) {Nm}^{2} / C ϕ = 0.24 \times 10^{3} {Nm}^{2} / C ϕ = 240 {Nm}^{2} / C$

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The electric field in a region is given by E→=35E0 i→+45 E0 j→ with E0=2·0×103 N C-1. Find the flux of this field through a rectangular surface of area 0⋅2 m2 parallel to the y-z plane.

The electric field in a region is given by $\vec{E} = \frac{3}{5} E_{0} \vec{i} + \frac{4}{5} E_{0} \vec{j} with E_{0} = 2 \cdot 0 \times 10^{3} N C^{- 1}$ . Find the flux of this field through a rectangular surface of area 0⋅2 m² parallel to the y-z plane.