- #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.
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.
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'?
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?
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??
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]
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.
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.
gabbagabbahey said:That seems like a more reasonable conclusion than "he is wrong"
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.
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.
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.
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.
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.