English

A Small Conducting Sphere of Radius 'R' Carrying a Charge +Q Is Surrounded by a Large Concentric Conducting Shell of Radius Ron Which a Charge +Q Is Placed. - Physics

Advertisements
Advertisements

Question

A small conducting sphere of radius 'r' carrying a charge +q is surrounded by a large concentric conducting shell of radius Ron which a charge +Q is placed. Using Gauss's law, derive the expressions for the electric field at a point 'x'
(i) between the sphere and the shell (r < x < R),
(ii) outside the spherical shell.

Advertisements

Solution 1

Consider a sphere of radius r with centre O surrounded by a large concentric conducting shell of radius R.

To calculate the electric field intensity at any point P, where OP = x, imagine a Gaussian surface with centre O and radius x, as shown in the figure given above.

The total electric flux through the Gaussian surface is given by

`ø = oint_s Eds = E oint_s \ ds`

Now,

`oint \ ds = 4πx^2`

`∴ø = E xx 4πx^2   ....... (1) `

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

`ø = q/ε_0    ....... (2)`

From (i) and (ii), we get

` E xx 4πx^2 = q/ε_0 `

⇒`E = q/(4πε_0x^2 )`

shaalaa.com

Solution 2

(1) Consider a sphere of radius r with centre O surrounded by a large concentric conducting shell of radius R.

To calculate the electric field intensity at any point P, where OP = x, imagine a Gaussian surface with centre O and radius x, as shown in the figure given above.

The total electric flux through the Gaussian surface is given by

`ø = oint_s Eds = E oint_s \ ds`

Now,

`oint \ ds = 4πx^2`

`∴ø = E xx 4πx^2   ....... (1) `

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

`ø = q/ε_0    ....... (2)`

From (i) and (ii), we get

` E xx 4πx^2 = q/ε_0 `

⇒`E = q/(4πε_0x^2 )`

(2)

To calculate the electric field intensity at any point P', where point P' lies outside of the spherical shell, imagine a Gaussian surface with centre O and radius x', as shown in the figure given above.
According to Gauss's theorem,

` E' xx 4πx^'2 = (q+Q)/ε_0 `

⇒`E = (q+Q)/(4πx'^ 2 )`

As the charge always resides only on the outer surface of a conduction shell, the charge flows essentially from the sphere to the shell when they are connected by a wire. It does not depend on the magnitude and sign of charge Q.

shaalaa.com
  Is there an error in this question or solution?
2014-2015 (March) Foreign Set 2

RELATED QUESTIONS

Find the electric field intensity due to a uniformly charged spherical shell at a point (ii) inside the shell. Plot the graph of electric field with distance from the centre of the shell.


A thin metallic spherical shell of radius R carries a charge Q on its surface. A point charge`Q/2` is placed at its centre C and an other charge +2Q is placed outside the shell at a distance x from the centre as shown in the figure. Find (i) the force on the charge at the centre of shell and at the point A, (ii) the electric flux through the shell.


Find the ratio of the potential differences that must be applied across the parallel and series combination of two capacitors C1 and C2 with their capacitances in the ratio 1 : 2 so that the energy stored in the two cases becomes the same.


An infinitely large thin plane sheet has a uniform surface charge density +σ. Obtain the expression for the amount of work done in bringing a point charge q from infinity to a point, distant r, in front of the charged plane sheet. 


A point object is placed on the principal axis of a convex spherical surface of radius of curvature R, which separates the two media of refractive indices n1 and n2 (n2 > n1). Draw the ray diagram and deduce the relation between the object distance (u), image distance (v) and the radius of curvature (R) for refraction to take place at the convex spherical surface from rarer to denser medium.


A spherical shell made of plastic, contains a charge Q distributed uniformly over its surface. What is the electric field inside the shell? If the shell is hammered to deshape it, without altering the charge, will the field inside be changed? What happens if the shell is made of a metal?


A rubber balloon is given a charge Q distributed uniformly over its surface. Is the field inside the balloon zero everywhere if the balloon does not have a spherical surface?


A thin, metallic spherical shell contains a charge Q on it. A point charge q is placed at the centre of the shell and another charge q1 is placed outside it as shown in the  following figure . All the three charges are positive. The force on the charge at the centre is ____________.


A circular wire-loop of radius a carries a total charge Q distributed uniformly over its length. A small length dL of the wire is cut off. Find the electric field at the centre due to the remaining wire.


If one penetrates a uniformly charged spherical cloud, electric field strength ______.

Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×