# Electric flux - Area vector question

1. Jul 18, 2012

### 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!

2. Jul 18, 2012

### 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.

Last edited: Jul 18, 2012
3. Jul 18, 2012

### Quine!

http://www.technology2skill.com/science_mathematics/vector_analysis/vector_picture/curve_surface_integral_first.png [Broken]

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!)

Last edited by a moderator: May 6, 2017
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