Question

find the flux of the constant vector field v→=3i→ 2j→−5k→ through a square plate of area 25 in the xz-plane oriented in the positive y-direction.

Answer 1

The flux across a surface is an indicator of the rate at which fluid is flowing through it when there is a fluid flowing in three dimensions and a surface positioned inside that space.

What is Electric Flux?

• A feature of an electric field is electric flux. One way to conceptualize it is as the quantity of forces interacting in a specific space.
• According to conventional wisdom, electric field lines begin with positive electric charges and conclude with negative charges.
• A closed surface's field lines are seen as negative when they are directed into it, and as positive when they are directed out of it.
• Every field line that is directed into a closed surface continues through the interior and is often directed outward somewhere else on the surface if there is no net charge present within the surface.
• The net, or total, electric flux is zero because the size of the negative and positive fluxes are just about equal.
• A closed surface that contains a net charge will have a total flux that is proportional to the enclosed charge, positive if it is positive and negative if it is negative.

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Answer 2

Answer:

The flux of the field through a square of 25 area whose plane is the xz-plane oriented in the positive y-direction.is 50

What is Flux?

Any effect that seems to pass through or move through a surface or substance is referred to as a flux, whether it actually flows or not. There are numerous applications of the concept of flux in applied mathematics and vector calculus. Flux, a vector quantity that describes the size and direction of the flow of a substance or attribute for transport phenomena. Flux is a scalar number in vector calculus, defined as the surface integral of a vector field's perpendicular component over a surface.

According to question

The flux of an field is given by,

ΦE = E.A = EA cos θ

where θ is the angle formed by the field lines and the normal (perpendicular) to the surface, S, and E, the magnitude of the field

vector field is v→=3i→ 2j→−5k→

Solving for each direction separately

Clearly

For i and k then flux is Zero because the angle b/w field and normal to surface is 90° and Cos(90) = 0

Now for j

E = 2

S = 25

θ = 0° (cos0° = 1)

Φ = EScosθ

    = 2*25*1

    = 50

Therefore,  the flux of the field through a square of 25 area whose plane is the xz-plane oriented in the positive y-direction.is 50

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