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.

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.

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.