# Why does the superposition principle work in method of images?

Okay, let there be a conducting sphere having radius $a$ initially charged with $Q$ & insulated. Now, $q$ is brought in front of the conductor at $y$ from the center.

Now, Jackson in his book uses the superposition principle to calculate the field & potential :

If we wish to consider the problem of an insulated conducting sphere with total charge $Q$ in the presence of a point charge $q$, we can build up the solution for the potential by linear superposition. In an operational sense, we can imagine with a grounded conducting sphere(with its charge $q'$ distributed over its surface). We then disconnect the ground wire & add to the sphere an amount of charge $(Q - q')$. ... .To find the potential, we merely note that the added charge $(Q - q')$ will distribute itself uniformly over the surface since the electrostatic forces due to point charge $q$ are already balanced by the charge $q'$.[...]

Okay, this seems to me correct since the force imparted by $q'$ on $\left(Q- q'\right)$ is cancelled by the field of $q$ since it is a zero equipotential surface of $q-q'$ system.

But why doesn't $\left(Q- q'\right)$ exert force on $q'$ on the surface?? On the other hand, superposition principle tells that you can't alter the configuration of either system when superimposing the two systems which means the surface charge density of $q'$ remains the same even when there are $\left(Q- q'\right)$ other charges on the surface?? So, why doesn't $\left(Q- q'\right)$ exert force on $q'$?? If it exerts force on $q'$, then it would change the distribution of $q'$ which can't happen as said by superposition principle. So, can anyone explain why $\left(Q- q'\right)$ wouldn't exert force on $q'$??

Okay, this seems to me correct since the force imparted by $q'$ on $\left(Q- q'\right)$ is cancelled by the field of $q$ since it is a zero equipotential surface of $q-q'$ system.

I have no idea why you think a zero potential surface has anything to so with anything. While the field of $q$ and the field of $q'$ together make a zero potential surface on the surface of sphere they most definitely exert net force on the charge $Q-q'.$

But why doesn't $\left(Q- q'\right)$ exert force on $q'$ on the surface??

It does. The charges that make up the density of charge $Q-q'$ exerts forces on each other so they exert force on the charges that make up the density of charge $q'$ in exactly the same way.

On the other hand, superposition principle tells that you can't alter the configuration of either system when superimposing the two systems which means the surface charge density of $q'$ remains the same even when there are $\left(Q- q'\right)$ other charges on the surface??

Let's be straight, when you grounded the sphere the external charge $q$ felt a force as if there were a charge $q'$ at the image charge location. And there is a net charge of $q'$ on the surface and the charges that make up the net charge of $q'$ exert forces on each other but the forces on each other add up to zero. But they still feel a force due to the charge $q$ and in fact the total force on all the charge on the surface is equal and opposite to the force the charge $q$ feels due to an image charge at the image location. And the actual force at each point in the sphere points orthogonal to the surface.

That orthogonality is the sole fact we get from the fact that the surface is an equipotential surface. And the force that is orthogonal is the total force i.e. the force due to the other charges on the surface plus the force due to the charge $q$ ... only their sum is orthogonal.

As for superposition, the force due to the uniform spread of charge $Q-q'$ over the surface is a force that points radially outwards as well.

The densities add, the fields add and the forces add. The only reason things don't move is that forces orthogonal the surface don't allow charges to move.

So, why doesn't $\left(Q- q'\right)$ exert force on $q'$??

It does.

If it exerts force on $q'$, then it would change the distribution of $q'$

No it doesn't, because the net force is orthogonal to the surface. And charges are free to move inside a conductor but they are not free to move outside a conductor. So a force orthogonal to the surface simply doesn't move the charges because the conductor itself is capable of keeping charges from leaving the conductor. It is called a work function.

which can't happen as said by superposition principle.

And now I think you misunderstand the superposition principle. It just says that if you add fields and add charges that it is also a solutions. And it does mean that if you add static solutions, you get static solution. If 100% everything were electrostatics then yes you could add like that. But consider the simple case of adding charge $q''$ uniformly spread to the outside of a conducting sphere, no grounds, no external charge. If you kept adding solutions like that over and over again eventually the work function might not be enough to keep such a huge charge density on the surface. Or maybe the field density outside the sphere gets large enough to break down the dielectric that is the air itself outside the sphere.

Just because you add charge and current and fields together doesn't mean that things don't move just because they didn't move on their own. Because the sphere itself or the air itself has to deal with all those things added together and it could have limits to what it can handle.

Superposition is just having a linear equation, add the sources and add the individual solutions and you get a solution. Anything non-linear such as a work function or a dielectric breakdown means the effects might not be static even if the individual things were static. Superposition doesn't mean more than it means.

Why did then Jackson write that the force on $Q - q'$ from $q$ is balanced by $q'$?

The forces due to $q$ and $q'$ are already balanced in the sense that the total field due to both is orthogonal to the surface so the new charge also arranges itself so that its field is also orthogonal to the surface as well which means the new charge is arranged uniformly.

• Firstly, thanks for the answer..... Just read the first para; why wouldn't zero potential mean no net force?? Let two equal but opposite charges be taken; suppose a test charge is just in front of the negative charge & hence it must be in negative potential. As it goes towards the positive charge, attractive force from '-' charge decreases & repulsive force from $+$ charge increases; eventually at the mid-point of the configuration, both attractive & repulsive force would become equal & opposite to each other. The potential there would be zero. Why? Because the potential is changing from ... – user36790 Aug 25 '15 at 15:15
• being negative at the negative charge region to being positive at the positive charge site. So, the region where the force from the opposite charges cancel each other is the zero equipotential region. Hence zero equi-potential region of any charge-configuration is the region where there is no net force. Am I wrong, sir? Isn't my argument in the example above, right?? – user36790 Aug 25 '15 at 15:17
• @user36790 Don't generalize from a single example. If you have equal and opposite charges at a fixed distance away from each other then the plane that bisects the line segment between them is the zero of potential and you identified the one point on that entire surface where the force is zero (and actually it's not zero there either you made an error). At every single other point the force isn't zero. And potential is related to energy I don't know why you would think energy is related to force in any way except that the gradient of the potential (energy) is proportional to the field (force). – Timaeus Aug 25 '15 at 15:23
• I want to know the error, please. And energy is not related to force?? As you wrote later, it is the negative of the gradient of the potantial. – user36790 Aug 25 '15 at 15:28
• @user36790 Yes. Though I thought we covered that earlier in a different question. Like physics.stackexchange.com/q/201718 – Timaeus Aug 26 '15 at 16:13

The problems solved by the method of images are, at their heart, boundary value problem involving the divergence of a scalar field, and those problem have uniqueness theorems that says there is only one configuration of fields in the volume that generates a particular set of boundary conditions.

So if you find any method of generating a field that meets the boundary value conditions, you can use the fields from that method as the field in the volume.

The method of images say "imagine that the actual physical situation was replaced with this charge distribution" that has the desired boundary values. And then uses the imaginary charge distributions to compute the field, and every computation of field from a charge distribution relies on the the superposition of field contributions.

• The image problem doesn't have a discontinuous electric field at the conductor surface so I don't think it directly tells you the force the surface charges feel from each other. It gives you the field outside the conductor but the field inside matters too. You could try to argue that the average of the two is related to the force but this doesn't follow from a simple uniqueness argument for a boundary value problem. – Timaeus Aug 26 '15 at 5:07
• Hmmm ... it seems I misunderstood the question. – dmckee Aug 26 '15 at 5:33
• You answered the title well, the body of the question basically asks how superposition can be consistent with the forces involved. – Timaeus Aug 26 '15 at 12:07