Applications of Gauss's Law in Integral Form
Explanation1
Gauss’s Law 1(../635) allows us to calculate the electric field very easily, but not always. Although Gauss’s Law itself always holds, its formulaic advantage can only be utilized in certain situations. As will be explained below, to easily calculate the electric field through Gauss’s Law, the magnitude of the electric field $\mathbf{E}$ must be constant, and its direction perpendicular, over the surface created by a specific coordinate system. In a spherical coordinate system, this means we can easily calculate the electric field that is uniform in magnitude and exits perpendicular to the spherical surface. Surfaces of constant electric field magnitude are called Gaussian Surfaces. There are three cases where a Gaussian surface can be drawn.
Gauss’s Law
$$ \oint_{\mathcal{S}} \mathbf{E} \cdot d\mathbf{a} = \frac{1}{\epsilon_{0}}Q_{\mathrm{in}} $$
Spherical Symmetry
Draw a spherical shell of radius $r>R$. This shell is the Gaussian surface. Since the direction of $d\mathbf{a}$ and $\mathbf{E}$ are the same, their dot product is simply the product of their magnitudes. Furthermore, as the electric field’s magnitude is constant on the spherical surface, it is a constant. Therefore, the Gauss’s Law formula simplifies as follows.
$$ \int \mathbf{E} \cdot d\mathbf{a} = \int \left|\mathbf{E} \right| da = \left|\mathbf{E} \right| \int da=\left|\mathbf{E} \right| 4\pi r^2 $$
By Gauss’s Law, since $\displaystyle \int \mathbf{E} \cdot d\mathbf{a}=\frac{1} {\epsilon_{0}} Q_\mathrm{in} = \frac{1}{\epsilon_{0}}q$, solving concurrently yields the following.
$$ \begin{align*} && \left|\mathbf{E} \right| 4 \pi r^2 &= \frac{1}{\epsilon_{0}}q \\ \implies && \left|\mathbf{E} \right| &= \frac{1}{4\pi \epsilon_{0}}\frac{q}{r^2} \\ \implies && \mathbf{E} &= \frac{1}{4\pi \epsilon_{0}}\frac{q}{r^2} \hat{\mathbf{r}} \end{align*} $$
The important point here is that measuring the electric field outside the sphere yields results as if the total charge $Q_\mathrm{in}$ were concentrated at the center of the sphere. In other words, how the charge $q$ is distributed inside the sphere is irrelevant.
Cylindrical Symmetry
Draw a cylinder of radius $s$ and length $l$ around the axis of a wire. This cylinder is the Gaussian surface. The electric field can be calculated in the same way as in Spherical Symmetry. Assuming the charge density is proportional to the distance from the wire, then $\rho=ks$ (where $k$ is a constant) and calculating $Q_{\mathrm{in}}$ yields the following.
$$ \begin{align*} Q_\mathrm{in}&=\int \rho d\tau \\ &= \int_{0}^l \int_{0}^{2\pi} \int_{0}^s (ks^{\prime})(s^{\prime}ds^{\prime}d\phi dz) \\ &= \frac{2}{3}k\pi l s^3 \end{align*} $$
Calculating the integration of the electric field gives the following.
$$ \int \mathbf{E} \cdot d\mathbf{a} = \int \left| \mathbf{E} \right| da = \left| \mathbf{E} \right| \int da = \left|\mathbf{E} \right| 2 \pi s l $$
It might seem odd that the surface area of the cylinder doesn’t include the ends. This is because at the ends, the direction of the electric field and the surface vector are perpendicular, resulting in a flux of $0$. It’s important to note that only the surfaces through which the electric field passes perpendicularly should be considered in the surface area calculation. Combining the two formulas gives $\displaystyle \left| \mathbf{E} \right| 2 \pi s l = \frac{1}{\epsilon_{0}}\frac{2}{3}k\pi l s^3$, resulting in the following.
$$ \mathbf{E} = \frac{1}{3\epsilon_{0}}ks^2 \hat{\mathbf{s}} $$
This calculation was made assuming the wire is infinitely long, but if the wire is long and sufficiently far from the ends, it can be considered an approximation.
Planar Symmetry
The solution follows the same logic. Draw two rectangular prisms with the same thickness above and below a plane. These two rectangular prisms are the Gaussian surfaces. Let’s say the area of the top (bottom) surface of the Gaussian surface is $A$. And if the surface charge density of the plane is $\sigma$, then the total charge enclosed by the Gaussian surface is as follows.
$$ Q_\mathrm{in}=\sigma A $$
Calculating the integration of the electric field gives the following.
$$ \int \mathbf{E} \cdot d\mathbf{a}=\left| \mathbf{E} \right| \int d\mathbf{a} = \left| \mathbf{E} \right| 2A $$
Therefore, by Gauss’s Law, the following holds.
$$ \int \mathbf{E} \cdot d\mathbf{a} = \frac{1}{\epsilon_{0}}Q_\mathrm{in}=\frac{1}{\epsilon_{0}}\sigma A $$
Since the flux on the side surfaces is $0$, only the flux through the top and bottom surfaces remains, yielding a value of $2A$. The reason why the flux on the side surfaces is $0$ is because the surface vector $d\mathbf{a}$ and the electric field $\mathbf{E}$ are perpendicular, resulting in a dot product of $0$. Summarizing yields the following.
$$ \mathbf{ E} = \frac{\sigma}{2\epsilon_{0}}\hat{\mathbf{n}} $$
Here, the important point is that regardless of the distance from an infinite plane, the magnitude of the electric field is the same. This is because the farther away from the plane, the more electric field lines reach that point. Think of it as being able to see more broadly the higher you go. Although moving away weakens the electric field, the influence from a larger area compensates for this, effectively canceling out the difference. Similar to Cylindrical Symmetry, if the plane is wide and far enough from the edges, it can be considered an approximation.
Conclusion
The electric field created by a charged sphere is proportional to $\dfrac{1}{r^2}$, the electric field created by an infinite wire is proportional to $\dfrac{1}{r}$, and the electric field created by an infinite plane is independent of distance.