In class, I argued that for two (conducting) charged plates with respective charge densities ±Q/A we...
In class, I argued that for two (conducting) charged plates with respective charge densities ±Q/A we can add their fields to find the overall field strength due to two parallel charged plates. While that is the correct result I want you to show that this indeed the case by considering the following scenario with conducting sheets of finite thickness:
(a) Take one negatively charged plate and add an uncharged conducting slab of equal dimension across from it such that they are parallel. Do you get any induced charges on it? Use Gauss’ law to argue where they must go. Comment on the overall field and sketch it as well as the charge distribution.
(b) Consider the previous scenario, but now add +Q to the previously neutral slab. Use Gauss’ law to determine the final charge distribution and field (sketch both).
Hint: Consider Gaussian surfaces that are cylinders, some of which have an end cap within the conductors. Also, ignore any edge effects due to the finite area of the plates.
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