हिंदी

State and Explain Gauss’S Law.

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
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
2014-2015 (March) Panchkula Set 3

संबंधित प्रश्न

A thin conducting spherical shell of radius R has charge Q spread uniformly over its surface. Using Gauss’s law, derive an expression for an electric field at a point outside the shell.


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


State Gauss's law in electrostatics. Show, with the help of a suitable example along with the figure, that the outward flux due to a point charge 'q'. in vacuum within a closed surface, is independent of its size or shape and is given by `q/ε_0`


The Electric flux through the surface


(i)

(ii)

(iii)

(iv)

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.

If there were only one type of charge in the universe, then ______.

  1. `oint_s` E.dS ≠ 0 on any surface.
  2. `oint_s` E.dS = 0 if the charge is outside the surface.
  3. `oint_s` E.dS could not be defined.
  4. `oint_s` E.dS = `q/ε_0` if charges of magnitude q were inside the surface.

In 1959 Lyttleton and Bondi suggested that the expansion of the Universe could be explained if matter carried a net charge. Suppose that the Universe is made up of hydrogen atoms with a number density N, which is maintained a constant. Let the charge on the proton be: ep = – (1 + y)e where e is the electronic charge.

  1. Find the critical value of y such that expansion may start.
  2. Show that the velocity of expansion is proportional to the distance from the centre.

In finding the electric field using Gauss law the formula `|vec"E"| = "q"_"enc"/(epsilon_0|"A"|)` is applicable. In the formula ε0 is permittivity of free space, A is the area of Gaussian surface and qenc is charge enclosed by the Gaussian surface. This equation can be used in which of the following situation?


A charge Q is placed at the centre of a cube. The electric flux through one of its faces is ______.


A charge of +5 μC is placed at the centre of two concentric spheres of radii r1 = 3 cm and r2 = 5 cm. The ratio of the flux through sphere of radius r1 to that through sphere of radius r2 will be ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×