Electric flux - Area vector question

okami11408

This is my first post here.

I have a problem dealing with Electric flux.

According to Gauss's Law, to find electric flux we use equation:

ψ=∫D dot dA, both D(Electric flux density) and dA(super small area normal to the surface) are vector quantity.

Now this is what I confused, how can "dA" be a vector quantity since it's an area.

I may misunderstand something.

Thank you!

Philip Wood

The electric flux [itex]d\Phi[/itex] through an area dA is defined as [tex]d\Phi=D\normalsize{d}A cos \phi[/tex].
[itex]\phi[/itex] is the angle between the normal to the area and [itex]\textbf{D}[/itex].
[itex]D cos \phi[/itex] is the component of D at right angles to the area.

We can write [itex]d\Phi=D\normalsize{d}A cos \phi[/itex] more neatly, as
[tex]d\Phi=\textbf{D}.d\textbf{A}[/tex]
if we define a vector [itex]d\textbf{A}[/itex] such that [itex]d\textbf{A}=\textbf{n}dA[/itex].

Here, [itex]\textbf{n}[/itex] is a unit vector normal to the area. If the area is part of a closed surface, we choose the normal pointing outwards from the enclosed volume.

Hope this helps. Congratulations on your first post. It is very clear.

Quine!

This might be a helpful picture: Here k is a flux vector, and n is the unit vector normal to the surface.

The reason why dA is a vector quantity is the amount of electric flux moving through a given Gaussian surface depends not only on the magnitude of the electric flux, but also its direction. For example, if the E-field is flowing directly perpendicular to the gaussian surface, there can't be any flow through the surface, and the flux will be zero. (You can see why the dot product/cos is used here!)