2
$\begingroup$

How is it with the electrostatic induction measurement? Everywhere you can read about the effect of electrostatic induction but I still can't found any detailed explanations or quantitative measurements. Perhaps someone here might help me.

Here is a Picture about the effect I am refering to:

Charging by induction

The process is realy simple. In the last picture the sphere is charged (one positive, one negative). But how high is the amount of electric charge? One can break this down to a more simple experiment: I have a capacitor which generates an electric field. I can calculate the homogeneously electric field by the voltage and dimension of the capacitor. Now I am moving two connected plates into this field, disconnect them and put them out of the field. One plate is charged positive, one negative. But how high is the amount of electric charge in dependence of the electric field?

I thought it might work by calculating the Electric displacement field? But I am willing to see Solutions by other People. Is it calculated this way or is there another method?

$\endgroup$
1
  • $\begingroup$ One would have to calculate the charge distribution in the geometry of the experiment, which, unfortunately, is usually only possible with numerical methods. There is no simple formula for how charges move on conductors because the charges cause an electric field and they also move in that field, i.e. one has to calculate the equilibrium solution. This is equivalent to calculating the coefficients of potential: en.wikipedia.org/wiki/Coefficients_of_potential. There may be a set of solutions for special geometries, like a set of spheres, though, but I am not aware of it, if it exits. $\endgroup$
    – CuriousOne
    Jul 5, 2016 at 8:33

1 Answer 1

0
$\begingroup$

Based of the shape of the conductors and the charge distribution on one body, it is theoretically possible to find the distribution on another body using the method of images. But this may lead to difficult mathematical problems(unsolvable in some cases).

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.