## Electric flux - Area vector question

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!

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 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.
 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!)

 Tags area vector question