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enter image description here

For example, if we had to find the field at Q1 due to the induced charges on the outside spherical surface ,what should be our approach?

According to me, as the induced charges due to q2 on the conducting sphere would be non uniformly distributed, the field inside the sphere itself will keep varying and hence it would be very difficult to calculate it. The induced charges due to Q1 on the conducting sphere will also create a problem as Q1 is not at the centre of the sphere and hence again the induced charges will be non uniformly distributed.

This question was asked an test and its answer was given as $\frac{q_1 q_2}{8\pi\epsilon_0 R^2}$ .

P.S I apologize for having such a small picture, I wasn't able to edit it properly due to some technical problems.

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  • $\begingroup$ Use the method of images to find the charge distribution inside. Then it is simple integration. $\endgroup$ – Lelouch Jul 9 '16 at 15:25
  • $\begingroup$ could you please elaborate a bit more on the method of images ? $\endgroup$ – Ishita Gupta Jul 9 '16 at 15:33
  • $\begingroup$ You wouldfind a better explanation in The book on electrodynamics by David J. Griffiths, or maybe even online. $\endgroup$ – Lelouch Jul 9 '16 at 15:44
  • $\begingroup$ @sammygerbil It is a wonderful answer, I was near to start to calculate some optimal function. But I am now very curious, how to calculate the virtual image charge. $\endgroup$ – peterh - Reinstate Monica Jul 9 '16 at 20:08
  • $\begingroup$ Could you clarify the question? You say you want the field at the point $q_1$ due induced charges caused by the outside charge, but in your discussion you mention the induced charges due to the inside charge. Is there a charge at location $q_1$ or not? If so, you might want to edit the subject and text to make that clear ... but it makes the problem as stated ambiguous. Also: do you know how far $q_2$ is from the center of the sphere? Perhaps you should write the statement of the problem exactly as given on the test. $\endgroup$ – garyp Jul 12 '16 at 13:59
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There seems to be something wrong with the answer.

My explanation is intuitive and quite long but worth it

Charges outside spherical conductor cannot induce electric field inside a electrically isolated conductor due to electrostatic shielding

Electrostatic Shielding-a region enclosed by meat of a conductor will not experience any electric field generated by charges placed outside the enclosure

So the field at the center is only due to the charges inside the shell and the inner surface.

NOTE:HERE WE CONSIDER $q_{2}$ IS NOT PRESENT

To understand this better consider a spherical hollow shell with a charge $q_{1}$ kept anywhere in it.we know that that this charge $q_{1}$ will induce a $-q_{1}$ on the inner surface of the shell and a charge $q_{1}$ on the outer surface of the shell.

On the inner surface of the shell the region close to the charge will have more induced charge density and region farther away from the charge will have less induced charge density.

The distribution on the induced charge on the outer surface of the shell is going to be spherically symmetric because there is no force acting on it,there is no electric field inside the meat of and electrically isolated conductor and outer induced charges are inside the conductor and will arranged themselves is such a way that the have minimum potential energy among them (due to repulsion),that is in a spherically symmetric manner

To find electric field at the center of the shell

  • Due to the charge induced on the inner surface of the shell
    region of high charge density are closer to $q_{1}$ but far away from center which balances the fraction $\frac{q}{r}\propto field$
    region of low charge density are far away from $q_{1}$ but closer to center which balances the fraction $\frac{q}{r}\propto field$
    So in net effect all the charges on the inner surface of the shell will produce (approx.)the same field at the center and will produce zero electric field.

  • Due to the charge induced on the outer surface of the shell Outer charges are symmetrically distributed and will not produce any field inside the shell hence field due to outer charge zero.

    • Due to $q_{1}$
      $$\frac{1}{4\pi\epsilon_{0}}\frac{q}{({\frac{r}{2}})^{2}}$$

Net electric field =(electric field due to the charge induced on the outer surface of the shell) + (electric field due to the charge induced on the outer surface of the shell)+(due to q1)
=0+0+$\frac{1}{4\pi\epsilon_{0}}\frac{q}{({\frac{r}{2}})^{2}}$ = $\frac{1}{4\pi\epsilon_{0}}\frac{q}{({\frac{r}{2}})^{2}}$

Now comes the most beautiful part

Consider a charge $q_{2}$ to be placed outside the shell.

It's effect on the outer surface of shell.
- it messes up the charge distribution

It's effect on the inner surface of shell.
- None.As long as there is the meat of the conductor protecting it experiences no electric field by $q_{2}$ hence no force by $q_{2}$(electrostatic shielding)

It's effect on electric field at the center
None. Surrounded by meat of the conductor - electrostatic shielding.

effect of the messed up charged distribution on the outside surface of the conductor on the electric field inside the conductor
None.electrostatic shielding

So electric field at the center
$$\frac{1}{4\pi\epsilon_{0}}\frac{q}{({\frac{r}{2}})^{2}}$$


Correction to this answers are welcomed


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  • $\begingroup$ Please do not post complete solutions to homework-like questions. Our policy on this can be found here which includes: “If someone posts an answer to a homework-type question that gives away a complete or near-complete solution, in most cases it will be temporarily deleted.” Please consider deleting this answer yourself, and instead comment on the problems and ambiguities in the statement of the problem. $\endgroup$ – garyp Jul 12 '16 at 14:02
  • $\begingroup$ The given answer is correct @garyp discussion $\endgroup$ – Koolman May 8 '17 at 3:11
  • $\begingroup$ physics.qandaexchange.com/?qa=2101/… $\endgroup$ – Koolman May 8 '17 at 3:13

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