The process of the method of images

AI Thread Summary
The discussion centers on the method of images in electrostatics, particularly regarding the grounding of an infinite conducting plane in relation to a positive charge. The main confusion arises from whether the plane should be grounded before or after placing the positive charge nearby. It is clarified that grounding the plane after placing the charge will neutralize the induced positive charges, leaving only the induced negative charges. Conversely, grounding the plane first and then introducing the positive charge will also induce negative charges on the plane, but the grounding does not prevent this induction. The conversation suggests that the principles of electrostatic shielding can help understand these scenarios, and an electroscope can be used for practical experimentation.
ta66505477
Messages
9
Reaction score
0
I am really confused that the whole process of the method of image.

Suppose I have the + q near the infinite conducting plane which is grounded.
My first question is: Is the conducting plane grounded before putting the +q or after putting the +q then grounded?

The first question would lead to my second question.

If I have an infinite neutral plane, then I am putting a +q near it, it will induce the negative charge on the plane. Next, I grounded the plane, so the positive charges on the plane will be neutralized (All the positive charges flowing away). After that, what you remain is the induced negative charges on the plane.

In another case, I am going to ground the infinite neutral plane first, then put the positive charge near the plane. Can the plane also be induced negative charges? If yes, will the +charges flow away?
 
Physics news on Phys.org
No matter you ground it first or later, I think the plane would both remain negative charges, which is induced by +q.
The method of the image can be grabbed directly by the concept of electrostatic shielding. Maybe you can try on it.
 
tommyxu3 said:
No matter you ground it first or later, I think the plane would both remain negative charges, which is induced by +q.
The method of the image can be grabbed directly by the concept of electrostatic shielding. Maybe you can try on it.
If you could find a gold leave electroscope you could try these experiments very easily.
 
Thank you!
I am going to find this electroscope.
 
Thread 'Gauss' law seems to imply instantaneous electric field propagation'
Imagine a charged sphere at the origin connected through an open switch to a vertical grounded wire. We wish to find an expression for the horizontal component of the electric field at a distance ##\mathbf{r}## from the sphere as it discharges. By using the Lorenz gauge condition: $$\nabla \cdot \mathbf{A} + \frac{1}{c^2}\frac{\partial \phi}{\partial t}=0\tag{1}$$ we find the following retarded solutions to the Maxwell equations If we assume that...
Maxwell’s equations imply the following wave equation for the electric field $$\nabla^2\mathbf{E}-\frac{1}{c^2}\frac{\partial^2\mathbf{E}}{\partial t^2} = \frac{1}{\varepsilon_0}\nabla\rho+\mu_0\frac{\partial\mathbf J}{\partial t}.\tag{1}$$ I wonder if eqn.##(1)## can be split into the following transverse part $$\nabla^2\mathbf{E}_T-\frac{1}{c^2}\frac{\partial^2\mathbf{E}_T}{\partial t^2} = \mu_0\frac{\partial\mathbf{J}_T}{\partial t}\tag{2}$$ and longitudinal part...
Thread 'Recovering Hamilton's Equations from Poisson brackets'
The issue : Let me start by copying and pasting the relevant passage from the text, thanks to modern day methods of computing. The trouble is, in equation (4.79), it completely ignores the partial derivative of ##q_i## with respect to time, i.e. it puts ##\partial q_i/\partial t=0##. But ##q_i## is a dynamical variable of ##t##, or ##q_i(t)##. In the derivation of Hamilton's equations from the Hamiltonian, viz. ##H = p_i \dot q_i-L##, nowhere did we assume that ##\partial q_i/\partial...
Back
Top