How to obtain the electric field and potential due to a fixed point charge inside a metal spherical shell?

Question

A simple solution is welcome.The shell will be at a fixed potential at all pts on its surface, I feel . Then how to reach the solution?
The charge particle is not at the centre and hence symmetry will not be available,but fields should be inside and outside the shell, I suppose. The distance of the ch is x from the centre of the shell x

Answer 1

There will be a surface charge induced on the sphere but it's not necessary to calculate it explicitly to get the fields. We can assume the total net charge on the sphere is zero. It's only necessary to find a solution which satisfies the boundary condition V(r=R)=Constant, and (for the outside field only) the additional condition V(r-->oo) = 0.

First the easier part: What's the potential outside the sphere? V(r=R)=constant determines the solution to be Constant/r, with the constant determined by Gauss's law. That is (using Gaussian units):

V = q/r

Of course the (radial) electric field is E=q/r^2.

The fields are the same as if the charge were located at the center of the sphere regardless of its actual location (as long as it's somewhere *inside* the sphere).

The field *inside* of the sphere is determined by the method of images. Choose a coordinate system so that the charge q is located on the z axis at (0,0,b), b < R.

The potential created by the surface charge is the same as what would be created by an image charge q' = -(R/b)q located at (0,0,R^2/b) (i.e. instead of a conducting sphere):

V = q/sqrt((z-b)^2+x^2+y^2) + q'/sqrt((z-b')^2+x^2+y^2)

where q' = -Rq/b and b' = R^2/b.

This can be verified by demonstrating that V is indeed constant (=0) everywhere on the surface x^2+y^2+z^2 = R^2:

V(anywhere on sphere)
= q/sqrt(R^2+b^2-2bz) + q'/sqrt(R^2+b^2-2zb')
= q/sqrt(R^2+b^2-2bz) + q/sqrt(b^2+R^2-2zb)
= 0

V is only defined to within an additive constant. The electric field inside the sphere is the gradient of V, in other words the same as the electric field due to two points charges q at (0,0,b) and q' and (0,0,b'):

E_x = qx/((z-b)^2+x^2+y^2)^(3/2) + q'x/((z-b')^2+x^2+y^2)^(3/2)
etc..

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Send me an e-mail if you don't think this answer is complete, or if something requires clarification.

Answer 2

A simple solution is to apply Gauss's law.

Suppose a fixed positive charge is placed inside the spherical cell (need not be at the center)

Using Gauss law we can say that this positive charge induces an equal amount of negative charge on the inner surface of the spherical cell and equal amount of positive charges on the outer surface of the spherical cell.

{Since this is a metal surface, free electrons are attracted by the positive charge and are on the inner surface and hence the outer surface acquire equal positive charge}

As the electrons pull the positive charge which is inside the spherical cell in all directions the charge must now come to the center of the spherical cell.

As for as the negative charges which are on the inner surface are concerned, using Gauss law we can treat these charges as if they are concentrated at the center of the sphere where there is already equal positive charge.

Hence net charge inside is zero.

In result, placing a positive charge inside a metal hollow sphere is the same as the charges are on the outer surface of the hollow sphere. Note that there are positive charges on the outer surface by induction.

Hence the field inside the hollow sphere is zero. That implies the potential is the same throughout the space inside the sphere.

For outside the sphere, the field can be calculated again by Gauss law, as if the charges are at the center of the hollow sphere.

Answer 3

The electic field (a vector) is given be the unfavorable of the gradient of the aptitude (a scalar). (this can provide electric powered field=replace in ability/distance for a uniform field, not electric powered field=replace in potentia*distance as in the previous answer) besides we are able to upload an arbitary consistent to the aptitude through the indisputable fact that is in ordinary words transformations in the aptitude that count number (it really is the electric powered field it quite is the actual volume that outcomes the action of debris and so on..) and so the aptitude ought to take any cost in the middle. in the middle of the sector the aptitude will seem somewhat like the bottom of a quadratic and so it really is gradient and for this reason the electric powered field will disappear.

Answer 4

if you would like to have a field inside the sphere place any sphere of which is carrying certain amount of charge inside it ;
i.e., radius of second sphere is less than the radius of the first sphere .it is so simple