मराठी

Obtain the expression for the electric field intensity due to a uniformly charged spherical shell of radius R at a point distant r from the centre of the shell outside it. - Physics

Advertisements
Advertisements

प्रश्न

  1. Obtain the expression for the electric field intensity due to a uniformly charged spherical shell of radius R at a point distant r from the centre of the shell outside it.
  2. Draw a graph showing the variation of electric field intensity E with r, for r > R and r < R.
आलेख
बेरीज

उत्तर

Electric field due to a uniformly charged thin spherical shell:


i. When point P lies outside the spherical shell: Suppose that we have calculate the field at point P at a distance r(r > R) from its centre. Draw the Gaussian surface through point P so as to enclose the charged spherical shell. The gaussian surface is a spherical surface of radius r and centre O.

Let `vecE` be the electric field at point P, then the electric flux through the area element of area `vec(ds)` is given by dφ = `vecE.vec(ds)`

Since `vec(ds)` is also along normal to the surface dφ = E dS

∴ Total electric flux through the Gaussian surface is given by

φ = `oint  Eds = E point  ds`

Now, `oint` ds = 4πr2  ...(i)

= E × 4πr2  

Since the charge enclosed by the Gaussian surface is q, according to Gauss’s theorem,

φ = `q/∈_0`  ...(ii)

From equations (i) and (ii) we obtain

E × 4πr2 = `q/∈_0`

E = `1/(4π∈_0) . q/r^2`  ...(For r > R)

ii. A graph showing the variation of electric field as a function of r is shown below.

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
2022-2023 (March) Sample

APPEARS IN

व्हिडिओ ट्यूटोरियलVIEW ALL [4]

संबंधित प्रश्‍न

A 36 cm long sonometer wire vibrates with frequency of 280 Hz in fundamental mode, when it is under tension of 24.5 N. Calculate linear density of the material of wire.


"For any charge configuration, equipotential surface through a point is normal to the electric field." Justify.


Obtain the formula for the electric field due to a long thin wire of uniform linear charge density λ without using Gauss’s law. [Hint: Use Coulomb’s law directly and evaluate the necessary integral.]


A point charge q is at a distance of d/2 directly above the centre of a square of side d, as shown the figure. Use Gauss' law to obtain the expression for the electric flux through the square.


If the point charge is now moved to a distance 'd' from the centre of the square and the side of the square is doubled, explain how the electric flux will be affected.


The electric field intensity outside the charged conducting sphere of radius ‘R’, placed in a medium of permittivity ∈ at a distance ‘r’ from the centre of the sphere in terms of surface charge density σ is


Charge motion within the Gaussian surface gives changing physical quantity ______.

A spherical ball contracts in volume by 0.02% when subjected to a pressure of 100 atmosphere. Assuming one atmosphere = 105 Nm−2, the bulk modulus of the material of the ball is:


Sketch the electric field lines for a uniformly charged hollow cylinder shown in figure.


Draw a graph of kinetic energy as a function of linear charge density λ.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×