6

I am a bit confused about when to use the method of images in E&M? For example, in Griffiths' Introduction to Electrodynamics, Example 3.2, the problem reads:

A point charge $q$ is situated a distance $a$ from the center of a grounded conducting sphere of radius $R$. Find the potential outside the sphere.

But in Example 2.7, which says,

Find the potential of a uniformly charged spherical shell of radius $R$

or Problem 3.1, which says,

Find the average potential over a spherical surface of radius $R$ due to a point $q$ located inside

the method of image is not used to solve the problem. In Example 2.7, this equation for the potential is used instead: $$ V\left(\mathbf r\right)=\frac{1}{4\pi\epsilon_0}\int\frac{\sigma}{r}da' $$

Is it because the object is "conducting" that we use the method of images? And if so, why? Could someone help me clear up my misunderstanding?

Nikhil Anand
  • 265
Guest
  • 375

4 Answers4

14

For problem solving, the method of images comes in handy in the following cases: conducting sphere, conducting cylinder, conducting ellipsoid, and conducting plane. Another example is two regions of dielectrics, with different $\epsilon$ (permittivity). And that is it. This general rule should save you some time.

So, the first problem you mentioned can be solved with images, the second and third cannot (there is no conductor).

A caveat about using the method of images, which I have seen frequently confuse students is as follows. In the method of images, there is always two distinct regions of space. For example, inside and outside the conductor, left and right of the conducting plane, or in dielectric region 1 and dielectric region 2. The imperative point is that, when you want to find the potential in one region, the image charges are not allowed to be in that region; they should be situated in the other region. For example, if there is a conducting sphere, not grounded, and there is a point charge inside of it (not at the center), and you are asked to find the potential for every point inside the sphere, then you should situate the image charge outside the sphere.

I suggest not confusing yourself too much with the method of images, and remembering that only the above-mentioned geometries have neat closed-form image solutions.

philophilosophia
  • 191
1

The method of images is useful when you have a point charge near a $V=0$, a.k.a. grounded, surface; usually a plane or a sphere. In this configuration you can substitute the surface by an additional point charge (of opposite charge) and the problem becomes finding the potential due to two point charges. In these other problems, you don't have this configuration. The method of images is useful, but very limited. It is called the method of images because the $V=0$ surface works as a mirror for the point charge.

Physico
  • 133
0

I think you are a little confused about what problem does the method solve. If you have a known distribution of charges $\rho$, then the potential $\phi$ is simply given by an integral: $$\phi = \dfrac{1}{4\pi \varepsilon _0} \int \frac{\rho \text d V}{r}.$$ This is the integral you have to solve to determine the potential in example 2.7..

Now, if you have a know distribution plus a conductor body, in the static situation the electric field must be normal on the conductor's surface i.e. this must be an equipotential surface. This doesn't give directly the surface charge density on the conductor, so you can't simply write an integral as in the previous case. All you have is the equation $\nabla ^2 \phi = -\rho/\varepsilon _0$, with boundary conditions $\phi = \phi _i$ on the $i-\text {th}$ surface. The method of images is a way to solve this problem, mathematically grounded on the fact that if the conductor's surfaces bound a volume, then the solution of the above equation is unique. If you can “invent” an image-distribution $\rho '$ inside the conductors, which together with $\rho$ gives the correct values on the surfaces, then the potential of this imaginary distribution $\rho +\rho '$, outside of the conductors, satisfies $\nabla ^2 \phi = -\rho /\varepsilon$, has the correct values on the surfaces and is therefore the solution.

pppqqq
  • 4,824
0

to use Gauss Law, you should have a system which is symmetric. I mean there should not be any boundary conditions and for such systems Gauss Law can be used in any problem. But the problems which you should use image methods or any further like Green's formula etc., are not symmetric. A point charge near a conducting sphere break the symmetry of electric field.

aQuestion
  • 1,151