English

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

Question

  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.
Graph
Sum

Solution

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
  Is there an error in this question or solution?
2022-2023 (March) Sample

APPEARS IN

RELATED QUESTIONS

Explain why, for a charge configuration, the equipotential surface through a point is normal to the electric field at that point


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


A point charge of 2.0 μC is at the centre of a cubic Gaussian surface 9.0 cm on edge. What is the net electric flux through the surface?


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.


Draw a graph to show the variation of E with perpendicular distance r from the line of charge.


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


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.


A solid metal sphere of radius R having charge q is enclosed inside the concentric spherical shell of inner radius a and outer radius b as shown in the figure. The approximate variation of the electric field `vecE` as a function of distance r from centre O is given by ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×