Induced charg by chagr distribution

AI Thread Summary
Calculating the induced charge on a conductor from a charge distribution, such as a conducting disc, can be approached using the method of image charges, which allows for the superposition of individual charge effects. While this method is effective for point charges, it can be extended to charge distributions by integrating over the distribution to determine the total field. The problem becomes more complex when the charge distribution on the conductor changes due to interactions with the induced charges, necessitating a different approach. By treating the conductors as equipotential surfaces and applying Laplace's equation with appropriate boundary conditions, one can find the new charge distributions. Analytical solutions may be available for some cases, but numerical methods may be required for more complex scenarios.
gabrielbhl
Messages
13
Reaction score
0
Hi everybody
It's easy (and well known) to calculate the charge distribution of a condutor in the presence of a point chage. BUT if we change this point charge and put an charge distribution... like a condutor disc, how can I calculate the induced chage of the other conductor??

THANKS
 
Physics news on Phys.org
Even using the method of image charges is is too dificul to calculate... I have no idea
 
If you have a geometry where you can use image charges then nothing is changed by replacing the point charge with a distribution of charges. Electromagnetics follows linear superposition and since this is a static problem your total solution will be the solution for each individual charge in your distribution. So you can just find the image of the charge distribution and calculate the field from the image.

Of course using images is only limited to a small subset of problems, but the point still stands. If you can solve for a single charge then you can just extend the method for a distribution of charges. The primary difference will lie in that you will have to integrate over the charge distribution (much like you would sum up over a finite set of charges) to get the total field.

This is assuming that your charge distribution will not change in the presence of an applied field. But if you were to charge a conducting disc then the charge distribution on your disc will change due to the interaction of the original charge distribution with the induced charge distribution on your conductor. In this case it is a much more difficult problem. However, you can solve this problem more easily by first noting that a conductor is an equipotential surface. So instead of trying to work from the standpoint of:
I have charge distribution A on a conducting object. I bring in conducting object B. What is the new charge distributions on A and B?

You work from the viewpoint of:
I have an object A with a potential of X. I bring in B. What is the charge distribution on A and B where A has a potential of X and B has a potential of 0 when they are infinitely separated?

You can represent this problem as a differential equation, Laplace's equation, and using the appropriate boundary conditions you can then solve for the potential on the surfaces of the objects. There are some problems where you can solve this analytically (method of images is just an example of this) but sometimes you need to use numerical techniques like method of moments or relaxation methods.
 
man.. thanks a lot, really, i was freaking out here... I needed to hear (read) it from someone. Thanks!
 
Consider an extremely long and perfectly calibrated scale. A car with a mass of 1000 kg is placed on it, and the scale registers this weight accurately. Now, suppose the car begins to move, reaching very high speeds. Neglecting air resistance and rolling friction, if the car attains, for example, a velocity of 500 km/h, will the scale still indicate a weight corresponding to 1000 kg, or will the measured value decrease as a result of the motion? In a second scenario, imagine a person with a...
Scalar and vector potentials in Coulomb gauge Assume Coulomb gauge so that $$\nabla \cdot \mathbf{A}=0.\tag{1}$$ The scalar potential ##\phi## is described by Poisson's equation $$\nabla^2 \phi = -\frac{\rho}{\varepsilon_0}\tag{2}$$ which has the instantaneous general solution given by $$\phi(\mathbf{r},t)=\frac{1}{4\pi\varepsilon_0}\int \frac{\rho(\mathbf{r}',t)}{|\mathbf{r}-\mathbf{r}'|}d^3r'.\tag{3}$$ In Coulomb gauge the vector potential ##\mathbf{A}## is given by...
Dear all, in an encounter of an infamous claim by Gerlich and Tscheuschner that the Greenhouse effect is inconsistent with the 2nd law of thermodynamics I came to a simple thought experiment which I wanted to share with you to check my understanding and brush up my knowledge. The thought experiment I tried to calculate through is as follows. I have a sphere (1) with radius ##r##, acting like a black body at a temperature of exactly ##T_1 = 500 K##. With Stefan-Boltzmann you can calculate...
Back
Top