A Doubt Regarding Quantization of charge ,its relevance in integration

[Uday]

a) We know that the smallest charge that can exist is 'e' . But in several instances (such as calculating potential energy of sphere of charge ) we consider 'dq' and then integrate it . How can we justify this ?

b) We know that 1/2 or 1/3 of e (charge of electron) doesn't exist . But integration is a process in which we add every fraction in the limits of integration which should be an error in the case of integration of charges . So should we use a different method for adding up ( in uniform distribution of charges ) ?
PLZ HELP ... and thanks for spending ur valuable time reading this ...

 

[phinds]

Sorry, I can't answer your question but at least I can point out to you that you do not have a "doubt", you have a "question":

https://www.physicsforums.com/showthread.php?t=607274

 

[WannabeNewton]

If we have a system consisting of $N \rightarrow \infty$ charges then we can treat the system as continuous to an extremely good approximation and perform integrals instead of extremely large sums. The error in the integral due to the discreteness of charge will be vanishingly small as $N \rightarrow \infty$. If you've taken statistical mechanics then you most likely used this approximation method over and over when calculating partition functions for various systems with many degrees of freedom. We're doing the same thing here.

 

[Uday]

If we have a system consisting of $N \rightarrow \infty$ charges then we can treat the system as continuous to an extremely good approximation and perform integrals instead of extremely large sums. The error in the integral due to the discreteness of charge will be vanishingly small as $N \rightarrow \infty$. If you've taken statistical mechanics then you most likely used this approximation method over and over when calculating partition functions for various systems with many degrees of freedom. We're doing the same thing here.

can we talk about dq charge?

 

[WannabeNewton]

Consider a system of $N$ charges, all of charge $q$, such that $N \gg 1$. If I add another charge $q$ to this collection then the new total charge will just be $Q = q(N + 1)$ but $N \gg 1$ so compared to $qN$ the new charge $q$ is so small, i.e. $q \ll qN$, that we can just treat $q$ as an infinitesimal amount of charge $dq$ and the error in this approximation will get smaller and smaller as $N \rightarrow \infty$.

So physically no there isn't $dq$ amount of a charge but if we have $N \rightarrow \infty$ amount of charges in a system then adding a new charge $q$ will add such a fantanstically small amount of charge to the total charge of the system that we may as well treat $q$ as an infinitesimal $dq$ and use integration, which is computationally much simpler to work with than sums of Coulomb forces between an extremely large number of charges.