1.14. APPLICATIONS OF GAUSS'S LAW

Consider an infinitely long thin straight wire with uniform linear charge density $\lambda$. The wire is obviously an axis of symmetry. Suppose we take the radial vector from O to P and rotate it around the wire. The points P, P¢, P¢¢ so obtained are completely equivalent with respect to the charged wire. This implies that the electric field must have the same magnitude at these points. The direction of electric field at every point must be radial (outward if $\lambda > 0$, inward if $\lambda < 0$). This is clear from Fig. 1.26.

Consider a pair of line elements P and P of the wire, as shown. The electric fields produced by the two elements of the pair when summed give a resultant electric field which is radial (the components normal to the radial vector cancel). This is true for any such pair and hence the total field at any point P is radial. Finally, since the wire is infinite, electric field does not depend on the position of P along the length of the wire. In short, the electric field is everywhere radial in the plane cutting the wire normally, and its magnitude depends only on the radial distance r.

To calculate the field, imagine a cylindrical Gaussian surface, as shown in the Fig. 1.26(b). Since the field is everywhere radial, flux through the two ends of the cylindrical Gaussian surface is zero. At the cylindrical part of the surface, E is normal to the surface at every point, and its magnitude is constant, since it depends only on r. The surface area of the curved part is $2 \pi r l$, where l is the length of the cylinder. Flux through the Gaussian surface = flux through the curved cylindrical part of the surface = $E \times 2 \pi r l$. The surface includes charge equal to $\lambda l$. Gauss’s law then gives $E \times 2 \pi r l = \lambda l/\epsilon_0$ $\Rightarrow E = \lambda/(2 \pi \epsilon_0 r)$. Vectorially, $\vec{E} = \frac{\lambda}{2 \pi \epsilon_0 r} \hat{n}$ (where $\hat{n}$ is the radial unit vector in the plane normal to the wire passing through the point). E is directed outward if $\lambda$ is positive and inward if $\lambda$ is negative.

In this chunk, we discuss the electric field generated by an infinitely long straight wire that is uniformly charged. The significant aspects include understanding the radial symmetry around the wire, which leads to the conclusion that the electric field's magnitude at any point around the wire is constant and only depends on the distance from the wire. This is determined using a cylindrical Gaussian surface to apply Gauss's law. The electric field is derived to be inversely proportional to the distance from the wire, mathematically represented as $E = \frac{\lambda}{2 \pi \epsilon_0 r}$. This shows that as you move further from the wire, the electric field strength decreases.

Imagine a long garden hose filled with water. If you consider the water flow to represent the electric field and the hose itself as the charged wire, the further away you stand from the hose, the less water (or electric field strength) you will feel. Just like the electric field that decreases as you move away from the charged wire, the water flow that you sense also decreases with distance.

Let $\sigma$ be the uniform surface charge density of an infinite plane sheet. We take the x-axis normal to the given plane. By symmetry, the electric field will not depend on y and z coordinates and its direction at every point must be parallel to the x-direction.

We can take the Gaussian surface to be a rectangular parallelepiped of cross-sectional area A, as shown. As seen from the figure, only the two faces 1 and 2 will contribute to the flux; electric field lines are parallel to the other faces and they, therefore, do not contribute to the total flux.

The unit vector normal to surface 1 is in –x direction while the unit vector normal to surface 2 is in the +x direction. Therefore, flux $\vec{E} \cdot \vec{D}S$ through both the surfaces is equal and adds up. Therefore the net flux through the Gaussian surface is $2EA$. The charge enclosed by the closed surface is $\sigma A$. Thus by Gauss’s law, $2E A = \sigma A/\epsilon_0$ or $E = \sigma / (2 \epsilon_0)$. Vectorically, $\vec{E} = \frac{\sigma}{2 \epsilon_0} \hat{n}$, where $\hat{n}$ is a unit vector normal to the plane and going away from it. E is directed away from the plate if $\sigma$ is positive and toward the plate if $\sigma$ is negative.

In this part, we analyze the electric field due to an infinite plane sheet with a uniform surface charge density. Because of the plane’s symmetry, the electric field it produces does not vary with position along the plane—it's constant. Using a rectangular Gaussian surface, we find that the net electric field is derived as $E = \frac{\sigma}{2 \epsilon_0}$. This result indicates that regardless of the distance from the plane, the electric field strength remains constant in magnitude and direction based on the charge density.

Think of a very large piece of paper with a uniform coating of electric paint. If you were to place a sensor directly at various distances away from the paper's surface, you will notice that the amount of 'electric paint' you measure would not change—this 'paint' represents the electric field, which stays consistent regardless of where you are relative to the surface.

Let $\sigma$ be the uniform surface charge density of a thin spherical shell of radius R. The situation has obvious spherical symmetry. The field at any point P, outside or inside, can depend only on r (the radial distance from the centre of the shell to the point) and must be radial (i.e., along the radius vector).

(i) Field outside the shell: Consider a point P outside the shell with radius vector r. To calculate E at P, we take the Gaussian surface to be a sphere of radius r and with center O, passing through P. All points on this sphere are equivalent relative to the given charged configuration. (That is what we mean by spherical symmetry.) The electric field at each point of the Gaussian surface, therefore, has the same magnitude E and is along the radius vector at each point. Thus, E and DS at every point are parallel and the flux through each element is E DS. Summing over all DS, the flux through the Gaussian surface is E × 4πr². The charge enclosed is $\sigma \times 4\pi R²$. By Gauss’s law $E \cdot 4\pi r² = \frac{q}{\epsilon_0}$.

(ii) Field inside the shell: In this case, the flux through the Gaussian surface, calculated as before, is $E \times 4 \pi r²$. However, in this case, the Gaussian surface encloses no charge. Gauss’s law then gives $E \cdot 4 \pi r² = 0$ i.e., E = 0 (r < R). This important result is a direct consequence of Gauss’s law which follows from Coulomb’s law. The experimental verification of this result confirms the 1/r² dependence in Coulomb’s law.

Here, we discuss the electric field resulting from a uniformly charged spherical shell. The spherical symmetry ensures that the electric field behaves uniformly outside the shell and is zero inside. Using Gauss's law, we calculate the external field to be $E = \frac{q}{4\pi \epsilon_0 r^2}$, mirroring the behavior of a point charge located at the center when outside the shell. However, for any point located inside the shell, the electric field is zero, reinforcing the concept that charge distribution does not affect the interior space of the shell.

Imagine a basketball inflated to full pressure, where the air inside represents the influence of the electrical field due to the charged shell. While the air pressure (electric field) is felt against the inner walls of the shell, inside the basketball itself, there's no pressure (electric field) felt if you were inside; the air (charge) only affects the outer part.

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