If a metallic sphere is grounded and close to a positive charge q, it will be charged with -q. Let's say that the electrons will arrive through the grounding. This charge will cover the surface of the sphere and counter the filed outside the sphere. This sphere will have a potential of zero with the ground as reference.

Now another scenario. If the sphere is not grounded then induced charge in the surface will counter the external field. Inside the sphere I read that a second field will be generated to nullify the field of the external charge. What will the potential of this sphere be?

If it is different than zero, why? Please give a qualitative answer and not a math one.


1 Answer 1


I'm assuming you mean a solid metallic sphere, not a shell. The analysis is slightly different for a shell but rests on the same principles. It is actually not correct that the total charge induced on the sphere in the case of grounding is $-q$. The basic concepts you need to know about electrostatics with conductors is that the entirety of the conductor must be an equipotential volume and that any net charge distribution must reside on the surface of the conductor. (Do you see why this must be the case? Remember that we are working in the statics case, so charges must not be accelerating and hence charges should experience no net force). Now, if you will forgive me, I will do some simplified math first, and then try to formulate the reasoning "qualitatively" as you ask.

Mathy (only slightly) version

This problem can be solved using the method of image charges, which basically works as follows: if we can put some additional "image" charges in the sphere, figure out the potential in all of space, and then restrict our solution to the outside, the solution we get is a solution the sphere for Maxwell's equation with just our original external charge $q$ (Maxwell's equations are local differential equations, and so if we have restricted ourselves to the region outside the sphere, we need only consider charge densities outside the sphere.) Thus, if we can choose our image charges so that the appropriate boundary conditions are satisfied, i.e., the surface of the sphere is an equipotential with appropriate potential, then the solution is correct for our original problem (this can be more rigorously formulated as a uniqueness theorem for solutions to the Poisson equation with Dirichlet boundary conditions).

Now suppose that the radius of the sphere is $R$ and the external charge $q$ is at a distance $d$ from the centre of the sphere. You can easily check (I will leave it up to you, if you'd like to, and if you need any help verifying this leave me a comment and I can edit my answer to show how), that if we place an image point charge of charge $-qR/d$ at a point in the sphere a distance $R^2/d$ from the centre of the sphere, on the line connecting the centre of the sphere to the external charge $q$, the potential on surface of the sphere (i.e. at $r=R$) vanishes. This corresponds, as you have noted, to the case of where the sphere is grounded. (Note that this is not to say that physically, there is a point charge that develops in the middle of the body of the metallic sphere - this is impossible in electrostatics. Rather, a surface charge density is created that from the outside behaves as if it were such a point charge. We could compute exactly what this surface charge density looks like fairly straightforwardly from the so called matching conditions at the interface between the sphere and the outside; again, if you'd like me to explain this to you, leave me a comment.)

The first thing to note is that in the grounded case, the total charge contained in the sphere is not $-q$, but rather $-qR/d$. You might feel that this is jumping the gun a bit: the image charge is an idealisation, not a physical charge density, and you might think that the physical surface charge density may have a different total charge. However, remember that whichever situation we examine, the solution for the potential outside the sphere is exactly the same, and therefore, the electric field outside the sphere is exactly the same. If we apply Gauss' law to a surface containing the metallic sphere, but not containing the external charges, the surface integral of the electric flux in the image charges picture and the physical surface charge distribution picture would have to be the same, since the electric fields are the same - but in both cases this relates directly to the enclosed charges, and so we can state generically that the total charge of the physical surface charge distribution must be the same as the total charge of the image charge in the distribution. In this case specifically, it must be $-qR/d$, not $-q$.

Now, coming to the second part of your question - what if we don't ground the sphere? Well, we would need to modify our image charge distribution somehow to account for the new boundary conditions. There are two conditions that we need to meet: first, the total charge in the image charge distribution must be zero, and secondly, the surface of the sphere must remain an equipotential surface (though possibly at a different potential). But by superposition, this is easy to do - just place a charge of $qR/d$ at the centre of the sphere! The total charge in the image charge distribution is now $-qR/d+qR/d=0$, and the second image charge at the centre simply changes the potential at the surface of the sphere from $0$ to $k(qR/d)/R=kq/d$. A curious point to note here: as long as $R<d$, i.e., the external charge is outside the sphere, the induced potential on the surface of the sphere doesn't actually depend on the radius of the sphere $R$ - and it is not at all obvious why this should be the case.

Qualitative version

Now, to come to the "qualitative" part of the answer. Can I give a simple, math-free explanation why the induced charge on the sphere is $-qR/d$ in the case of grounding? No, I can't (perhaps someone smarter could). But we can see that it could not be simply $-q$. If I had the external charge at a distance $d\gg R$ away from the centre of the sphere, i.e., very, very far away, the potential due to the external charge at all points on the surface of the sphere would be approximately zero, and so there would be no need for an induced surface charge density to correct the potential in the sphere. In fact, we can deduce from this argument that our result for the total charge density must go to $0$ as $d$ goes to infinity, which is the case for our answer.

Having accepted the answer to the first part, for the second part we are basically asking: ok I know how to distribute surfaces charges on the sphere so that the total potential in the sphere is $0$; but the total charge of these surface charges is $-qR/d\neq 0$. If I want to make the sphere neutral, I will need to distribute an additional charge of $qR/d$ over the surface of the sphere. Now, I am allowed to change the potential of the surface of the sphere from zero, but I must still respect the condition that the potential is constant on the sphere - thus, I must distribute the charge of $qR/d$ uniformly on the surface of the sphere, by symmetry considerations. Now, what is the potential due to this uniform surface charge distribution of $qR/d$? From the shell theorem (look it up if you are unfamiliar, it is a neat result, and fairly intuitive), since this charge distribution is spherically symmetric, outside of itself the potential due to this charge shell behaves as if all the charge was concentrated at a point at the centre and inside the charge shell, the potential is constant, i.e., outside of the sphere, for $r>R$, the potential due to this additional, uniform surface charge distribution of $qR/d$ is given by $kqR/dr$. Evaluating this at (technically, just outside) the surface of the sphere,we get $kq/d$, as before - and since the potential is continuous and constant inside the sphere it must be $kq/d$ everywhere in the sphere. Since the potential without this distribution was zero throughout thesphere, by the principle of superposition, the total potential at the surface of the sphere is $kq/d\neq 0$ throughout the sphere.

Even more qualitative version

Even more qualitatively, without assuming the result in the grounded case, we can make an argument for why the potential cannot be generically zero (though not much more). Assume WLOG that the point charge $q$ is positively charged, and suppose that the potential was generically zero throughout the sphere. It is intuitively obvious that the surface charge density induced on the sphere must be most negative at the point closest to the external charge (the "north pole") and least negative (equivalently most positive) at the diametrically opposite point, furthest from the external charge (the "south pole"), and that travelling along a line of "constant longitude" between the two points, the surface charge density must vary continuously, becoming less and less negative (equivalently, more an more positive).

Now, in order for the sphere to be neutral overall, at least some part of the surface must have positive surface charge density, and so from the argument above there exists some "latitude" above which the surface charge density is negative and below which it is positive. The total negative charge in the "northern" piece must equal the total positive charge in the "southern" piece for the entire sphere to remain neutral. Now, as we bring the external point charge closer and closer to the surface of the sphere, consider what happens to the potential at the "south pole".

There are three contributions to the potential - (1) a positive term from the eternal point charge, (2) a negative term from the negative "northern" piece and (3) a positive term from the positive "southern" piece. When we the external charge is very, very close to the surface of the sphere ($d\approx R$, but still $d>R$), as we are bringing it still closer, the term (1) does not change much - the distance to the south pole remains approximately the same at $\approx 2R$. However, as the charge gets closer and closer, negative charge gets pulled more and more towards the "north pole" and the charge in the "northern" piece gets increasingly concentrated in the north pole. The "southern" piece is not as strongly affected since all of the charges in it are further away from the external charge, but still, the positive charges are pushed more and more towards the "south pole". What happens to the relative sizes of term (2) and term (3)? Well, as we get closer, while the total charge in the "northern" piece may get more negative, and in the "souther " piece may get more positive, the total charges in both pieces remain equal to each other. But the negative charge of the "northern" piece is getting more and more concentrated towards the north pole, where as the positive charge in the "southern" piece is getting more and more concentrated towards the south pole.

Since we are interested in the potential at the south pole, we can see that the negative term (2) is becoming more negative at a slower rate than the rate at which positive term (3) is becoming more positive. This is obvious since the charges accumulating around the "south pole" are closer to the "south pole" than the charges accumulating around the "north pole", and hence give greater and greater contributions. Thus, it is clear that at least in the limit where we take the eternal charge very, very close to the sphere, since (1) remains fairly constant and (3) is increasing faster than (2) is decreasing, the potential at the south pole (and by extension, the potential on the whole sphere) must be increasing and thus cannot be generically zero.

  • $\begingroup$ Regarding the non-grounded case, there doesn't seem to be anything in the construction of the question/system whereby the total charge on the sphere has to be zero. The sum of the image charges which give a uniform potential on the surface must be zero, but any fixed charge could be added to the sphere (say by setting up the grounded situation, then cutting the connection to ground so the net charge is $-qR/d$). The scalar potential is gauge invariant, so absolutely any value of the potential is theoretically permissible in the non-grounded case. Or have I missed something? $\endgroup$
    – tok3rat0r
    May 13, 2015 at 22:26
  • $\begingroup$ @tok3rat0r We are just assuming a neutral sphere that happens to be close to an external charge. It should remain neutral. $\endgroup$ May 14, 2015 at 6:00
  • $\begingroup$ So, if I get this straight, there is a potential appearing in the non-grounded sphere since there is separation of positive and negative charge, something like a capacitance? Now I am stuck at how the potential can be everywhere (on the surface too) the same (in both cases - true in any case for conductors) when the charge is not uniformly distributed. Does this has to do with the fact that the electric field is normal to the surface? Also, can you please explain $\endgroup$ May 14, 2015 at 6:04
  • $\begingroup$ Also, I read that the R^2/d position is derived through similar triangles. Is this by chance or is there a reason? $\endgroup$ May 14, 2015 at 6:14
  • $\begingroup$ @NikolaosKakouros: You may be assuming a neutral sphere, but nowhere in the question do you state that it must be neutral. Regarding your second point, the potential is uniform because the displaced, negative image charge exactly cancels the field from the external positive charge, so the spatial potential variation cancels. The positive image charge which is added at the centre of the sphere just adds a fixed potential, uniform everywhere on the surface of the sphere. $\endgroup$
    – tok3rat0r
    May 14, 2015 at 10:58

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