1
$\begingroup$

Let us say we have the following uniformly and positively charged thin disk:

enter image description here

Suppose also that the total charge on the disk equas $Q$, with which we could define a constant charge density $\sigma$ equaling $\dfrac{Q}{\pi a^2}$ .

The charge $dq$ enclosed within the ring of thickness $dr$ may be computed in two ways:

  • Method 1: $$dq = \frac{dr}{a} Q$$

  • Method 2:$$dq=2\pi r dr \, \sigma=2rdr\frac{Q}{a^2}$$

Thus we can say that: $$\frac{dr}{a}Q=2rdr\frac{Q}{a^2}$$

$$r=\frac{a}{2} $$

The last result is incorrect since $r$ is a variable, are there any wrong assumptions that led to this inconsistency?

$\endgroup$
  • $\begingroup$ I don't understand where you got method 1. Can you elaborate on that? $\endgroup$ – Aaron Stevens Aug 9 at 23:35
  • $\begingroup$ It is realy just the percentage of the surface occupied by the ring, if $a$ corresponds to area of the full disk, then $dr / a$ will correspond to the area of the ring, we can also check the correctness of the formula by integrating the $dq$ given in method 1 over all the disk, i.e. r going from 0 to $a$. $\endgroup$ – Hilbert Aug 9 at 23:41
  • $\begingroup$ But $a$ isn't an area. It is the radius of the disk. $\text dr/a$ isn't an area, it is a dimensionless quantity $\endgroup$ – Aaron Stevens Aug 9 at 23:42
  • $\begingroup$ Thank you, indeed you are right, my first method does not take into consideration how far is the ring from the center, since the farther it is, the more charge it will contain, a fact which is not reflected by the first formula. $\endgroup$ – Hilbert Aug 9 at 23:47
1
$\begingroup$

So I think your method $1$ is obtained by assuming that each ring has a fraction $\text dr/a$ of the total charge on the disk (correct me if I am wrong about this). This is not correct. The amount of charge on a ring depends on how large the radius of that ring is, with rings with a larger radius having more charge. You cannot assume each ring holds an equal amount of charge.

Therefore, your method $2$ is correct since it correctly gives the dependence of the area of the ring on its radius.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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