## The Electric Field from a Point Charge

We use Gauss's law to determining the electric field of a point charge. Spherical symmetry is introduced to provide a deeper understanding of the physics. It includes interactive explanations, visualizations, and mathematics, where interaction with any of these components makes corresponding changes in the other components.

Both diagrams show the electric field from a point charge. The first diagram shows the physical charge along with the electric field lines radiating out from the charge. The grey surface is neutral and will be used to evaluate Gauss's law.

The second diagram shows the magnitude of the electric field vs $r$ , the distance from the charge.

Spin the field around in the first diagram. While individual field lines move, the field as a whole looks the same after any rotation in any direction. This is an example of spherical symmetry. Another way to visualize spherical symmetry is that the field lines are equally spaced, so the field has the same strength in every direction.

We quickly see two important things from the figure. The field lines are everywhere perpendicular to the neutral surface and they are evenly distributed around the surface. This makes the expressions in Gauss's law easier to evaluate. Surfaces where we evaluate Gauss's law are usually called Gaussian surfaces. One of the main motivations for selecting a specific Gaussian surface for a problem is that it simplifies the evaluation of Gauss's Law.

#### A Positive Charge Surrounded By A Gaussian Sphere

The field lines radiate out from the charge and pass through the sphere.

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### Gauss's Law

Gauss's Law relates the electric flux through a closed surface to the charge contained within that surface.

$${\iint}_{S}\mathbf{E}\cdot d\mathbf{A}=4\pi q$$We can evaluate this integral over the sphere centered on the charge to give the flux through the surface.

$${\iint}_{S}\mathbf{E}\cdot d\mathbf{A}={\iint}_{S}E\mathrm{cos}\left(\theta \right)dA\phantom{\rule{0ex}{0ex}}E\phantom{\rule{.5em}{0ex}}\text{is perpendicular to the surface.}={\iint}_{S}E\mathrm{cos}\left(0\right)dA={\iint}_{S}EdA\phantom{\rule{0ex}{0ex}}E\phantom{\rule{.5em}{0ex}}\text{is constant over the surface.}=E{\iint}_{S}dA=EA=E4\pi {r}^{2}$$Now that we know the flux through the surface, the next step is to find the charge contained within the surface. This will give us both sides of Gauss's law. In this case it is simply the point charge.

$$\phantom{\rule{5.3em}{0ex}}=4\pi q$$Setting these two sides of Gauss's law equal to one another gives for the electric field for a point charge:

$$E\phantom{\rule{4.3em}{0ex}}=\frac{q}{{r}^{2}}$$Then for our configuration, a sphere of radius $r=15.00\text{cm}$ centered around a charge of $q=150\text{statC}$ .

$$E\phantom{\rule{4.3em}{0ex}}=\frac{q}{{r}^{2}}=\frac{150\text{statC}}{{\left(15.00\text{cm}\right)}^{2}}=0.66\frac{\text{statV}}{\text{cm}}$$Next up, we will apply Gauss's law to the slightly more complex case of a line charge.