Gauss's Law | Physics for IIT JAM, UGC - NET, CSIR NET PDF Download

Gauss's law

The total of the electric flux out of a closed surface is equal to the charge enclosed divided by the permittivity.

Gauss`s Law | Physics for IIT JAM, UGC - NET, CSIR NET

Field Lines and Electric Flux

Consider that a point charge q is situated at the origin: Gauss`s Law | Physics for IIT JAM, UGC - NET, CSIR NET

This field is represented by the field line as shown in figure below. 

The magnitude of the field is indicated by the density of the field lines:

  •  It's strong near the center where the field line are close together, and 
  • Weak farther out, where they are relatively far apart. Gauss`s Law | Physics for IIT JAM, UGC - NET, CSIR NET

The field strength (E) is proportional to the number of field lines per unit area (area perpendicular to the lines).  

 The flux of E through a surface S, Gauss`s Law | Physics for IIT JAM, UGC - NET, CSIR NETis a measure of the “number of field lines” passing through S.Gauss`s Law | Physics for IIT JAM, UGC - NET, CSIR NET

For the case of point charge at the origin, the flux of Gauss`s Law | Physics for IIT JAM, UGC - NET, CSIR NET through a sphere of radius r is

Gauss`s Law | Physics for IIT JAM, UGC - NET, CSIR NET

Note that, any surface whatever its shape, would trap the same number of field lines. So the flux through any surface enclosing the charge is Gauss`s Law | Physics for IIT JAM, UGC - NET, CSIR NET 

Now suppose that instead of a single charge at the origin, we have a bunch of charges scattered about. 

According to the principle of superposition, the total field is simply the (vector) sum of all the individual fields: Gauss`s Law | Physics for IIT JAM, UGC - NET, CSIR NETThe flux through any surface that encloses them all, then, is Gauss`s Law | Physics for IIT JAM, UGC - NET, CSIR NETGauss`s Law | Physics for IIT JAM, UGC - NET, CSIR NET

A charge outside the surface would contribute nothing to the total flux, since its field lines go in one side and out the other. It follows, then, that for any closed surface,Gauss`s Law | Physics for IIT JAM, UGC - NET, CSIR NET

where Qenc is the total charge enclosed within the surface.   

This is Gauss’s law in integral form.

Question for Gauss's Law
Try yourself:
What does Gauss's law state about the total electric flux out of a closed surface?
View Solution

We can convert Gauss’s law in integral form to differential form, for continuous charge distributions, by applying the divergence theorem:Gauss`s Law | Physics for IIT JAM, UGC - NET, CSIR NET

We can write  Qenc in terms of the charge density Gauss`s Law | Physics for IIT JAM, UGC - NET, CSIR NET, we haveGauss`s Law | Physics for IIT JAM, UGC - NET, CSIR NET

So Gauss’s law becomesGauss`s Law | Physics for IIT JAM, UGC - NET, CSIR NET

Since this holds for any volume, the integrands must be equal: Gauss`s Law | Physics for IIT JAM, UGC - NET, CSIR NET

This is Gauss’s law in differential form.

Applications of Gauss’s Law 

Gauss's law is always true, but it is not always useful. Gauss's law is useful for only three kinds of symmetry:

  •  Spherical Symmetry: Make your Gaussian surface a concentric sphere. 

Gauss`s Law | Physics for IIT JAM, UGC - NET, CSIR NET

  • Cylindrical Symmetry: Make your Gaussian surface a coaxial cylinder. Gauss`s Law | Physics for IIT JAM, UGC - NET, CSIR NET
  • Plane Symmetry: Make your Gaussian surface a “pillbox,” which extends equally above and below the surface. Gauss`s Law | Physics for IIT JAM, UGC - NET, CSIR NET

If the charge distribution lacks sufficient symmetry for the application of Gauss' law, then the field must be found by summing the point charge fields of individual charge elements. 

Question for Gauss's Law
Try yourself:
In which of the following situations would Gauss's Law be most applicable for calculating the electric field?
View Solution

Example 1: Find the electric field inside and outside a spherical shell of radius R, which carries a uniform surface charge density Gauss`s Law | Physics for IIT JAM, UGC - NET, CSIR NET

Solution: Draw a spherical surface of radius r < R , which is called as “Gaussian surface”. According to Gauss’s Law  Gauss`s Law | Physics for IIT JAM, UGC - NET, CSIR NETGauss`s Law | Physics for IIT JAM, UGC - NET, CSIR NETGauss`s Law | Physics for IIT JAM, UGC - NET, CSIR NET

For outside point, draw a spherical surface of radius r > R  , 

According Gauss’s Law  Gauss`s Law | Physics for IIT JAM, UGC - NET, CSIR NET

Gauss`s Law | Physics for IIT JAM, UGC - NET, CSIR NETGauss`s Law | Physics for IIT JAM, UGC - NET, CSIR NET

Example 2: Find the electric field a distance r from an infinitely long straight wire, which carries a uniform line charge Gauss`s Law | Physics for IIT JAM, UGC - NET, CSIR NET

Solution: Draw a Gaussian cylinder of length l and radius r.  Gauss`s Law | Physics for IIT JAM, UGC - NET, CSIR NET

The document Gauss's Law | Physics for IIT JAM, UGC - NET, CSIR NET is a part of the Physics Course Physics for IIT JAM, UGC - NET, CSIR NET.
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FAQs on Gauss's Law - Physics for IIT JAM, UGC - NET, CSIR NET

1. What is Gauss's Law and how is it applied in electrostatics?
Ans.Gauss's Law states that the electric flux through a closed surface is proportional to the charge enclosed within that surface. It is expressed mathematically as Φ_E = Q_enc/ε_0, where Φ_E is the electric flux, Q_enc is the enclosed charge, and ε_0 is the permittivity of free space. In electrostatics, it is used to calculate electric fields around symmetric charge distributions, such as spherical, cylindrical, or planar charges.
2. How can Gauss's Law be used to determine the electric field of a charged sphere?
Ans.To determine the electric field of a charged sphere, we can apply Gauss's Law by considering a spherical Gaussian surface concentric with the charged sphere. For a uniformly charged sphere, the electric field outside the sphere behaves as if all the charge were concentrated at the center. Inside the sphere, the electric field is zero if the sphere is a conductor. The formula E = (1/4πε_0)(Q/r^2) can be derived for points outside the sphere, where E is the electric field, Q is the total charge, and r is the distance from the center.
3. In what scenarios is Gauss's Law particularly useful?
Ans.Gauss's Law is particularly useful in scenarios involving high symmetry, such as spherical, cylindrical, or planar charge distributions. It simplifies calculations for electric fields in cases where direct integration of Coulomb's Law would be complex or cumbersome. Examples include calculating the electric field of a charged infinite plane sheet, a charged cylinder, or a uniformly charged sphere.
4. Can Gauss's Law be applied to magnetic fields?
Ans.Yes, Gauss's Law can also be applied to magnetic fields, known as Gauss's Law for magnetism. This law states that the total magnetic flux through a closed surface is zero, indicating that there are no magnetic monopoles; instead, magnetic field lines are continuous. Mathematically, it is expressed as ∮ B · dA = 0, where B is the magnetic field and dA is the differential area vector. This principle is crucial in understanding the behavior of magnetic fields in various configurations.
5. What are some limitations of Gauss's Law?
Ans.Some limitations of Gauss's Law include its dependence on symmetry; it is most effective when dealing with symmetrical charge distributions. In cases where the charge distribution is irregular or lacks symmetry, applying Gauss's Law directly can be challenging. Additionally, while Gauss's Law provides a powerful tool for calculating electric fields, it does not directly give information about the potential energy associated with electric fields, which may require different approaches to fully analyze.
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