0
$\begingroup$

In my textbook, it is said: "Since $d$ is much smaller than the linear dimension of the plates ($d^2 << A$), we can use the result on electric field by an infinite plane sheet of uniform surface charge density." Why can't $d$ be large if we choose to use the result of electric field due to an infinite plane sheet in calculating the electric field by the plates of a parallel plate capacitor?

$\endgroup$
1
  • $\begingroup$ Of course it can be large. Usually d^2<<A if you want a decent amount of capacitance and a decent size of the capacitor at the same time. If you want to see the mathematics and derive some analytical expressions (and in textbook you usually do) in most cases you must have a simple model, and it "just happens" that electric field of a finite plate resembles that of an infinite plane at distances d^2<<A. $\endgroup$
    – Arturs C.
    Commented May 11, 2017 at 14:04

1 Answer 1

0
$\begingroup$

If it is not true that $d^2 << A$, then the approximations that were used in deriving the formula for capacitance are not valid, so the result is not valid.

You can make a capacitor with a larger $d$, but you will have to find some other way of calculating its capacitance.

$\endgroup$

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.