# What is \(\vec{E} \cdot \vec{dA}\) anyways?

**Electricity and Magnetism**Level 3

We now know how to integrate a function over a surface. The electric field is, however, a vector, and in order to integrate a vector over a surface we need a little more. One can't integrate a vector over a surface and get just a number using the method we previously described, as such a process would give a vector again. Since in Gauss' law the right hand side is just a number, just doing

\(\int_S \vec{E} dA\)

wouldn't work. Fortunately, there is a mathematical way to perform the correct integral. \(dA\) represents a small element of the surface \(S\) with area \(dA\). Think of this as a small square. Now if we put a square in 3-d space, there are two vectors of length one perpendicular to the square, where by perpendicular we mean that the angle between the vector and any line segment lying within the square is 90 degrees. Let \(\vec{n}_1\) and \(\vec{n}_2\) be these vectors. We can now define a vector version of \(dA\) via

\(d\vec{A}=\vec{n} dA\)

where \(\vec{n}\) is either \(\vec{n}_1\) or \(\vec{n}_2\).

Given such a vector, we can convert \(\vec{E}\) into a number by the familiar dot product, i.e. \(\vec{E}\cdot d\vec{A}\).

If \(\vec{E}=(x,y,z)\) what is \(\vec{E}\cdot \vec{n}\) evaluated on the surface of a sphere of radius 1? \(n\) is the outward pointing normal vector of the sphere.