Method of images in Griffiths' text

In summary, the conversation revolves around the derivation of equations (3.15) and (3.16) in Griffiths' text "Introduction to electrodynamics", specifically questioning their definition and the physical significance. The conversation also discusses a discrepancy in the calculation of work done using the method of images, with the conclusion that the author may have been considering the work done by the person bringing the charge into position rather than the work done by the field. There is also a mention of potential errors in Griffiths' text and the possibility of the author being wrong.
  • #1
KFC
488
4
In Griffiths' text "Introduction to electrodynamics", p.125, how does the author get (3.15) and (3.16)? I try for an hour to deduce the relation with geometry, but still can't get that.
 
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  • #2
KFC said:
In Griffiths' text "Introduction to electrodynamics", p.125, how does the author get (3.15) and (3.16)? I try for an hour to deduce the relation with geometry, but still can't get that.

in my version of Griffiths', (3.15) and (3.16) are definitions (of q' and b)... so, could you write down the question without reference to Griffiths'?
 
  • #3
olgranpappy said:
in my version of Griffiths', (3.15) and (3.16) are definitions (of q' and b)... so, could you write down the question without reference to Griffiths'?

Thanks for reply.

(3.15) is [tex]q' = -\frac{R}{a}q[/tex]

(3.16) is [tex]b = \frac{R^2}{a}[/tex]

I don't know why this is definition, I think we could deduce those expression with geometry; otherwise, what's the physical significance of defining q' like that?
 
  • #4
KFC said:
Thanks for reply.

(3.15) is [tex]q' = -\frac{R}{a}q[/tex]

(3.16) is [tex]b = \frac{R^2}{a}[/tex]

I don't know why this is definition, I think we could deduce those expression with geometry; otherwise, what's the physical significance of defining q' like that?

Right. So you want to know how to come up with those values. This is discussed by Griffiths in the chapter. To show that these values indeed do give a zero potential on the sphere is shown in problem 3.7, I think. Give it a try. Cheers.
 
  • #5
Thanks for reply. I will do it myself again.

Now, I just found another problem in using method of images in Grifiths' text. In page 124, there the author trying to calculate the work done by moving the charge from infinity to d and get the result

[tex]W=\int_\infty^d\vec{F}\cdot d l = \frac{1}{4\pi\epsilon_0}\int_\infty^d\frac{q^2}{4z^2}dz = \frac{1}{4\pi\epsilon_0}\left.\left(-\frac{q^2}{4z}\right)\right|_\infty^d = -\frac{1}{4\pi\epsilon_0}\frac{q^2}{4d}[/tex]

But according to the preceding calculation, the force on the charge due to the image charge is (eq. 3.12)

[tex]\vec{F} = -\frac{1}{4\pi\epsilon_0}\frac{q^2}{(2d)^2}\hat{z}[/tex]

The force is negative so it seems that in the work calculate we should plug the minus sign there, the result should be positive??
 
  • #6
KFC said:
Thanks for reply. I will do it myself again.

Now, I just found another problem in using method of images in Grifiths' text. In page 124, there the author trying to calculate the work done by moving the charge from infinity to d and get the result

[tex]W=\int_\infty^d\vec{F}\cdot d l = \frac{1}{4\pi\epsilon_0}\int_\infty^d\frac{q^2}{4z^2}dz = \frac{1}{4\pi\epsilon_0}\left.\left(-\frac{q^2}{4z}\right)\right|_\infty^d = -\frac{1}{4\pi\epsilon_0}\frac{q^2}{4d}[/tex]

But according to the preceding calculation, the force on the charge due to the image charge is (eq. 3.12)

[tex]\vec{F} = -\frac{1}{4\pi\epsilon_0}\frac{q^2}{(2d)^2}\hat{z}[/tex]

The force is negative so it seems that in the work calculate we should plug the minus sign there, the result should be positive??

Yes, since the force and the displacement are in the same direction the work should be positive. The 2nd equality after 'W' is wrong. The sign should be
[tex]
\frac{-1}{4\pi\epsilon_0}\int_{\infty}^{d}\frac{q^2}{4z^2}dz\;.
[/tex]

Maybe you got the sign wrong by double counting. E.g., [itex]d\vec\ell[/itex] should be [itex]\hat z dz[/itex] not [itex]-\hat z dz[/itex] because the direction is already accounted for in the limits of integration.

But the important point is that you recognized that the answer should be positive, which is right. So, however many minus sign errors you make you can always put the sign correct at the end.
 
  • #7
Thanks. But the following expression is given by Griffiths' text not me, that's why I am wondering the sign

[tex]W=\int_\infty^d\vec{F}\cdot d l = \frac{1}{4\pi\epsilon_0}\int_\infty^d\frac{q^2}{4z ^2}dz = \frac{1}{4\pi\epsilon_0}\left.\left(-\frac{q^2}{4z}\right)\right|_\infty^d = -\frac{1}{4\pi\epsilon_0}\frac{q^2}{4d}[/tex]
 
  • #8
KFC said:
Thanks. But the following expression is given by Griffiths' text not me, that's why I am wondering the sign

[tex]W=\int_\infty^d\vec{F}\cdot d l = \frac{1}{4\pi\epsilon_0}\int_\infty^d\frac{q^2}{4z ^2}dz = \frac{1}{4\pi\epsilon_0}\left.\left(-\frac{q^2}{4z}\right)\right|_\infty^d = -\frac{1}{4\pi\epsilon_0}\frac{q^2}{4d}[/tex]

So, either he is considering two charges of the same sign in which case the expression you gave for the force is wrong, or he is wrong.
 
  • #9
Or he is considering the work done by "you" in bringing the charge into position rather than the work done by the field.
 
  • #10
olgranpappy said:
Or he is considering the work done by "you" in bringing the charge into position rather than the work done by the field.

Oh. I see. How careless I am LOL
 
  • #11
olgranpappy said:
Or he is considering the work done by "you" in bringing the charge into position rather than the work done by the field.

That seems like a more reasonable conclusion than "he is wrong" :smile:
 
  • #12
gabbagabbahey said:
That seems like a more reasonable conclusion than "he is wrong" :smile:

not necessarily... Griffiths may write sacred texts, but he is far from infallible.
 

1. What is the method of images in Griffiths' text?

The method of images is a mathematical technique used in electrostatics to solve problems involving conductors and charges. It involves using imaginary charges, known as image charges, to simulate the behavior of a real charge in the presence of a conductor.

2. When is the method of images used?

The method of images is used in situations where a real charge is located near a conducting surface, such as a conducting plane, sphere, or cylinder. It is also used to calculate the electric field and potential of a point charge inside or outside a conducting sphere.

3. How does the method of images work?

The method of images works by applying the principle of superposition to the electric field and potential. The electric field and potential due to the real charge and the image charge are calculated separately, and then added together to obtain the total electric field and potential at any point.

4. What are the advantages of using the method of images?

One of the main advantages of the method of images is that it simplifies complex electrostatic problems involving conductors and charges. It also provides an intuitive understanding of the behavior of charges near conducting surfaces and allows for easier visualization of the electric field and potential.

5. Are there any limitations to the method of images?

The method of images is limited to problems involving electrostatics, as it does not take into account the effects of time-varying fields. It also requires a good understanding of the properties of conductors and the principles of electrostatics in order to accurately apply the method.

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