Suppose we have a capacitor charged with charge $Q1$

Then an electron beam hits the positive plate of the capacitor, thus increasing the charge of the capacitor (the battery will begin to deposit the electrons on the blue plate, thus increasing the charge such that $Q2> Q1$)

My question is, what is the maximum charge that a capacitor can support before it breaks?

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  • $\begingroup$ Why do you think the maximum charge achievable this way would be different from the maximum from just connecting a variable voltage source? $\endgroup$ – The Photon Aug 2 '20 at 15:52

That depends on the material used as a dielectric between the plates and the distance between them. Different materials are used as dielectrics depending on the capacitor and its intended use.

The dielectric material can be made from a number of insulating materials or combinations of these materials with the most common types used being: air, paper, polyester, polypropylene, Mylar, ceramic, glass, oil, or a variety of other $materials^1$.

The measure of how much the capacitance increases for the dielectric material is called the permittivity or dielectric constant.

Ref. 1 https://www.electronics-tutorials.ws/capacitor/cap_1.html

  • $\begingroup$ Does it also depend on the voltage of the battery source? $\endgroup$ – Danny LeBeau Aug 2 '20 at 16:54
  • $\begingroup$ Not is the sense of a property of the capacitor. The voltage is whats causing the breakdown! $\endgroup$ – Natsfan Aug 2 '20 at 16:55
  • $\begingroup$ My question is if the capacitor is already charged initially with a maximum initial charge Q1, but then these electrons enter the capacitor and the battery accommodates them in such a way that more charges accumulate, then Q2> Q1 my question is what is the maximum Q2 value, that's my question $\endgroup$ – Ricardo Casimiro Aug 2 '20 at 16:57

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