We know that the potential at the center of the charged sphere (if it is conductor) would be

$$V_o = \frac{1}{4\pi E_0}\frac{+q}{x}$$

If the distance between charges is x from their center of the sphere O

But then I was thinking that since if we take any positive charge near to the sphere then electric potential would be same but due to positive charge near sphere there will be induced charge on the sphere due to which near positive charge side there will be negative charge on the sphere as it will attract and just opposite to it there will be positive charge side as in figure.

Now since taking induced charge also in consideration then is there will be any change in the electric potential as there will electric field internally or it would be the same as I mentioned above ?

enter image description here

  • $\begingroup$ I'm a little unclear on what you're asking for. Let me try to rephrase it, and see if I've got it right: A conductor has zero net charge on it, and a point charge $q$ is placed a distance $x$ from its center. The electric potential of the conductor is $V = q/(4 \pi \epsilon_0 x)$. But this point charge will create positive and negative induced charges on the surface, which we have not taken into account. Do these surface charges create an electric field inside the conductor, and do they change the potential? $\endgroup$ – Michael Seifert May 14 '15 at 18:35
  • $\begingroup$ I should add that I can answer the question if it's phrased this way, I just want to make sure that it's the right question to answer. :-) $\endgroup$ – Michael Seifert May 14 '15 at 18:36
  • $\begingroup$ @MichaelSeifert yeah you are right that's my question $\endgroup$ – Shashank May 15 '15 at 1:24

Short answer: yes, the surface charges are taken into account; in fact, they're what ensures that $\vec{E} = 0$ inside the conductor.

The electric field at any point in space can be viewed as the superposition of the fields from the point charge outside the sphere, and the induced surface charges: $$ \vec{E} = \vec{E}_\text{point} + \vec{E}_\text{induced} $$ Now, inside the conductor, the electric field must be zero; the usual argument for this is that if the electric field wasn't zero inside the conductor, then the charges would move around in response to it and we wouldn't have a stable configuration. So as we bring the point charge in from infinity towards the conducting sphere, the positive and negative charges rearrange themselves to cancel out the field inside the conductor. In other words, for points inside the conductor, we must always have $$ \vec{E}_\text{induced} = - \vec{E}_\text{point}. $$ The potentials inside the sphere, too, must cancel out to within a constant (namely, the potential of the sphere: $$ V_\text{induced} = \frac{q}{4 \pi \epsilon_0 x} - V_\text{point}. $$

A cute side-effect of this phenomenon (credit to Bob Geroch for posing a similar problem to me years ago) is the following: Suppose we could somehow freeze the induced surface charge in place on the sphere, and then remove the point charge. The electric field inside the sphere would then look exactly like there was a negative point charge at the same location outside the sphere, like an "electric afterimage". The equipotentials inside the sphere would be concentric arcs, centered at a point outside the sphere:

enter image description here

(Apologies for the clunky field line diagram; Mathematica is not well-adapted to making field line diagrams. The field lines do not, of course, end anywhere except at the surface of the sphere.)

For points outside the sphere, of course, this cancellation of the electric fields doesn't take place, and the electric field is non-zero. However, it is still the case that the potential is constant over the outer surface of the sphere; it must be so, or the electric field wouldn't vanish inside the conductor. If the sphere was an insulator, then points on the side of the sphere facing the charge would be at higher potential, and points on the side of the sphere away from the charge would be at lower potential. From the above diagram, it's not too hard to see that the effect of the surface charges is to lower the potential at points on the sphere that would otherwise be at higher potential, and vice versa; the net effect is that the sphere is at constant potential, as desired.

  • $\begingroup$ Sorry, I couldn't understand. What will be the final potential at the center of the sphere and for a point outside it ? $\endgroup$ – Shashaank Sep 16 '17 at 23:37

You would simply add the potential that would exist at $O$ in the absence of the charge sphere and the potential which exists due to the sphere. This is due to superposition, since you can add the electric fields linearly and you must follow the same path in the path integral $V = -\oint \vec{E} \cdot \vec{dr}$ then the potentials actually add linearly as well.

  • $\begingroup$ So according to you there will be some difference ? $\endgroup$ – Shashank May 16 '15 at 1:44
  • $\begingroup$ Yes. A difference of kq/r where r is the distance from the new point charge to the center of the sphere. $\endgroup$ – JoDraX May 16 '15 at 1:45

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