Animated Solution for Physics - Electrostatics: The electric field in a region is given by $\mathbf{E}=\frac{2}{5}{E}_0\hat{\mathbf{i}}+\frac{3}{5}{E}_0\hat{\mathbf{j}}$ with $E_0=4.0\times 10^3\text{ N/C}$. The flux of this field through a rectangular surface area $0.4{\text{ m}}^2$ parallel to the yz-plane is ............ ${\text{N-m}}^2{\text{ C}}^{-1}$.

When Geometry Meets the Electric Field

Electric flux is one of those concepts that sounds abstract but is incredibly visual. Imagine standing in front of a fan; the amount of air hitting your face depends on how you angle your face. Electric flux is exactly the same, but instead of air, we have electric field lines, and instead of your face, we have a mathematical surface.

In this problem, we are given an electric field $\mathbf{E}=\frac{2}{5}{E}_0\hat{\mathbf{i}}+\frac{3}{5}{E}_0\hat{\mathbf{j}}$. Notice that this field has components in both the x and y directions. It's pointing diagonally across the xy-plane.

We are also given a rectangular surface with an area of $0.4{\text{ m}}^2$. The crucial piece of information is its orientation: it is parallel to the yz-plane.

To find the electric flux $\varphi$, we use the dot product of the electric field vector and the area vector:

But what is the area vector $\mathbf{A}$? By definition, an area vector is always perpendicular (normal) to the surface. Since our surface lies parallel to the yz-plane, the direction perpendicular to it is the x-axis. Therefore, the area vector points in the $\hat{\mathbf{i}}$ direction.

We can write our area vector as:

Now, we substitute our vectors into the flux equation:

When we perform the dot product, we multiply the corresponding components. The $\hat{\mathbf{j}}$ component of the electric field is parallel to the surface, meaning it skims right over it without piercing through. Mathematically, $\hat{\mathbf{j}}\cdot \hat{\mathbf{i}}=0$.

Only the $\hat{\mathbf{i}}$ component pierces the surface:

We are given that $E_0=4.0\times 10^3\text{ N/C}$. Let's plug this in:

Simplifying the numbers:

This problem beautifully illustrates why the dot product is so powerful in physics. It automatically filters out the parts of the field that don't contribute to the flux, leaving only the perpendicular component that truly matters!