मराठी

State and Explain Gauss’S Law. - Physics

Advertisements
Advertisements

प्रश्न

State and explain Gauss’s law.

Advertisements

उत्तर

Gauss’s law states that the flux of the electric field through any closed surface S is 1/∈ₒ times the total charge enclosed by S

Let the total flux through a sphere of radius r enclose a point charge q at its centre. Divide the sphere into a small area element as shown in the figure.

The flux through an area element ΔS is

`Deltaphi=E.DeltaS=q/(4piin_0r^2)hatr.DeltaS`

Here, we have used Coulomb’s law for the electric field due to a single charge q.

The unit vector `hatr`is along the radius vector from the centre to the area element. Because the normal to a sphere at every point is along the radius vector at that point, the area element ΔS and `hatr` have the same direction. Therefore

`Deltaphi=q/(4piin_0r^2)DeltaS`

Because the magnitude of the unit vector is 1, the total flux through the sphere is obtained by adding the flux through all the different area elements.

 `phi=sum_(all DeltaS)q/(4piin_0r^2)DeltaS`

 Because each area element of the sphere is at the same distance r from the charge,

`phi=q/(4piin_0r^2)sum_(all DeltaS)DeltaS=q/(4piin_0r^2)S`

Now, S the total area of the sphere equals 4πr². Thus,

`pi=q/(4piin_0r^2)xx4pir^2=q/in_0`

Hence, the above equation is a simple illustration of a general result of electrostatics called Gauss’s law

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
2014-2015 (March) Panchkula Set 3

संबंधित प्रश्‍न

A point charge +10 μC is a distance 5 cm directly above the centre of a square of side 10 cm, as shown in the Figure. What is the magnitude of the electric flux through the square? (Hint: Think of the square as one face of a cube with edge 10 cm.) 


Draw a graph of electric field E(r) with distance r from the centre of the shell for 0 ≤ r ≤ ∞.


Answer the following question.
State Gauss's law for magnetism. Explain its significance.


State Gauss’s law on electrostatics and drive expression for the electric field due to a long straight thin uniformly charged wire (linear charge density λ) at a point lying at a distance r from the wire.


q1, q2, q3 and q4 are point charges located at points as shown in the figure and S is a spherical gaussian surface of radius R. Which of the following is true according to the Gauss' law?


The surface considered for Gauss’s law is called ______.


Which of the following statements is not true about Gauss’s law?


If `oint_s` E.dS = 0 over a surface, then ______.

  1. the electric field inside the surface and on it is zero.
  2. the electric field inside the surface is necessarily uniform.
  3. the number of flux lines entering the surface must be equal to the number of flux lines leaving it.
  4. all charges must necessarily be outside the surface.

Refer to the arrangement of charges in figure and a Gaussian surface of radius R with Q at the centre. Then

  1. total flux through the surface of the sphere is `(-Q)/ε_0`.
  2. field on the surface of the sphere is `(-Q)/(4 piε_0 R^2)`.
  3. flux through the surface of sphere due to 5Q is zero.
  4. field on the surface of sphere due to –2Q is same everywhere.

If the total charge enclosed by a surface is zero, does it imply that the elecric field everywhere on the surface is zero? Conversely, if the electric field everywhere on a surface is zero, does it imply that net charge inside is zero.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×