Question

An electric field (2i + 3j ) x 103 nc-1 exists in a region. Calculate the electric flux linked with a square of side 0.5m held parallel to



a.Y-z plane



b.X-y plane.

Answer 1

Explanation:

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Answer 2

Concept Introduction: Vectors are used everywhere.

Given:

We have been Given: Electric Fields vector is

$(2i + 3j) \times 103 \: n {c}^{ - 1}$

Side of Square is

$0.5m$

To Find:

We have to Find: Electric Flux at

1. Y-Z plane.
2. X-Y plane.

Solution:

According to the problem,

The Flux in Y-Z plane is

$\beta = e.a \\ \beta = 2 \times 0.5 \times 0.5 = 0.5n {m}^{2} {c}^{ - 1}$

The Flux in the X-Y plane is

$\beta = 0$

Final Answer:

1. Y-Z plane flux is $0.5n {m}^{2} {c}^{ - 1}$
2. X-Y plane flux is $0$

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