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 d\Phi through an area dA is defined as d\Phi=D\normalsize{d}A cos \phi.
\phi is the angle between the normal to the area and \textbf{D}.
D cos \phi is the component of D at right angles to the area.

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

Here, \textbf{n} 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!]

http://www.technology2skill.com/science_mathematics/vector_analysis/vector_picture/curve_surface_integral_first.png 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!)