Gauss's Law

Gauss's Law

It relates the total flux (Φ) of an electric field (E) through a closed surface (S) to the net charge (q) enclosed by that surface. Let the electric field E makes an angle θ with the positive normal to the surface ΔS. Then, the quantity

ΔΦ = E ΔS cos θ is called the flux of the electric field through the chosen surface. If we draw a vector of magnitude ΔS along the positive normal, it is called the area vector, ΔS. Then

Note that the flux linked with a closed surface is independent of the shape and size of the surface and position of the charges inside it. While the field E itself depends on the mutual configuration of the charges, the flux of E through an arbitrary closed surface S is determined by the algebraic sum of charges inside S. If the charges insides are displaced, the field E may change considerably, but flux of E remains unchanged.

Note that the electric field E is the resultant field at the gaussian surface which also includes the contributions of charges outside the surface. 

In certain situations Gauss’s law can be used to find the electric field without using Coulomb’s law.

How to solve problems using Gauss’s Law?

(i)Use the symmetry of the charge distribution to determine the pattern of the field lines.

(ii)Choose a Gaussian surface for which E is either parallel to dS or perpendicular to dS.

(iii)If E is parallel to dS, then the magnitude E should be constant over this part of the surface. The integral then reduces to a sum over area elements.

Application of Gauss's Law 

Using Gauss's law and, in some cases, symmetry arguments, we an derive several important results in electrostatic situations. Among them are 

1.An excess charge on a conductor is located entirely on the outer surface of the conductor. 

2.The external electric field near the surface of a charged conductor is perpendicular to the surface and has magnitude 

E = σ/ε₀ (conducting surface) 

within the conductor, E = 0. 

3.The electric field at any point due to an infinite line of charge with uniform linear charge density λ is perpendicular to the line of charge and has magnitude

E = λ/2πε₀r (line of charge) 

where r is the perpendicular distance from the line of charge to the point. 

4.The electric field due to an infinite nonconducting sheet with uniform surface charge density σ is perpendicular to the plane of the sheet and has magnitude 

E = σ/2ε0 (sheet of charge)

5.The electric field outside a spherical shell of charge with radius R and total charge q is directed radially and has magnitude 

E = 1/4πε_0. q/r²(spherical shell, for r ≥ R). 

Here r is the distance from the center of the shell to the point at which E is measured. (The charge behaves, for external points, as if it were all located at the center of the sphere). The field inside a uniform spherical shell of charge is exactly zero. 

E = 0 (spherical shell, for r < R). 

6.The electric field inside a uniform sphere of charge is directed radially and has magnitude 

E = (q/4πε₀R³)r

Example 1.5

A charge of 4 × 10-8 C is distributed uniformly on the surface of a sphere of radius 1 cm. It is covered by a concentric, hollow conducting sphere of radius 5 cm. 

(a)Find the electric field at a point 2 cm away from the center. 

(b)A charge of 6 × 10-8 C is placed on the hollow sphere. Find the surface charge density on the outer surface of the hollow sphere.

Solution 

(a)Let us consider figure(a). Suppose, we have to find field at the point P. Draw a concentric spherical through P. All the points on this surface are equivalent and by symmetry, the field at all these points will be equal in magnitude and radial in direction. 

The flux through this surface =

where x = 2 cm = 2 × 10-2 m. 

From Gauss's law, this flux is equal to the charge q contained inside the surface divided by εo. Thus, 

4πx2E = q/εo

or  = 9 × 105 N/C 

(b)See figure (b). Take a Gaussian surface through the material of the hollow sphere. As the electric field in a conducting material is zero, the flux ∮EdS through this enclosed must be zero. Hence, the charge on the inner surface of the hollow sphere is -4 × 10-8 C. But the total charge given to this hollow sphere is 6 × 10-8 C. Hence, the charge on the outer surface will be 10 × 10-8 C. 

Example 1.6

Two conducting plates A and B are placed parallel to each other. A is given a charge Q1 and B a charge Q2. Find the distribution of charges on the four surfaces.

Solution 

Consider a Gaussian surface as shown in figure. Two faces of this closed surface lie completely inside the conductor where the electric field is zero. The flux through these faces is, therefore, zero. The other parts of the closed surface which are outside the conductor are parallel to the electric field and hence the flux on these parts is also zero. The total flux of the electric field through the closed surface is, therefore, zero. From Gauss's law, the total charge inside this closed surfaces of A should be equal and opposite to that on the inner surface of B. 

The distribution should be like the one shown in figure. 

To find the value of q, consider the field at a point P inside the plate A. Suppose, the surface area of the plate (one side) is A. Using the equation E = σ/(2εo), the electric field at P 

due to the charge Q1 - q =   (downward) 

due to the charge   (upward), 

due to the charge   (downward), 

and due to the charge  (upward).

The net electric field at P due to all the four charged surface is (in the downward direction). 

As the point P is inside the conductor, the field should be zero. Hence, 

Using these equations, the distribution shown in the figure can be redrawn as follows. 

This result is a special case of the following result. When charged conducting plates are placed parallel to each other, the two outermost surfaces get equal charges and the facing surfaces get equal and opposite charges.

Table 1.1 Electric field E due to Various Charge Distributions