1
$\begingroup$

I actually got a problem in understanding while doing physics. It is said that the maximum value of electric field on axis of a charged ring of radius a is at a distance $\frac{a}{\sqrt2}$. This corresponds to when $\frac{\mathrm{d}}{\mathrm{d}x}$ of expression for E = 0. What does $\frac{\mathrm{d}E}{\mathrm{d}x}=0$ mean exactly here? If rate of change of electric field with distance is zero, does it mean that if an object is placed at $\frac{a}{\sqrt2}$ and moved slightly by a distance $\mathrm{d}x$, the field remains the same? In that case when will it begin to change? Would someone please give an easy intuitive explanation of $\frac{\mathrm{d}}{\mathrm{d}x}$ being zero at a point?

$\endgroup$
  • 1
    $\begingroup$ You might want to take a look at critical points. $\endgroup$ – EuYu Aug 25 '18 at 11:28
1
$\begingroup$

Not sure about the problem but the strength of the electrical field, $E$, depends on your distance from it, which I assume is $x$. $$\frac{dE}{dx}$$ then, is how much the strength of the field changes as you move along the the axis. Now suppose that we're at a point, $P$, where $\frac{dE}{dx}=0$, what would happen to the field if me moved backwards, and what would happen if we moved forward? We got a lot of possibilities:

  • Nothing happens, $\frac{dE}{dx}=0$ when going forward and backward. This would mean that the field doesn't change at all, so $E$ must be constant.
  • When going backward $\frac{dE}{dx}<0$, that is $E$ is getting smaller, and when going forward $\frac{dE}{dx}<0$ also, that is $E$ is again getting smaller. So we're at a point $P$, and no matter which direction we move the field is getting smaller, this would mean that the field is strongest at $P$.
  • With the case $\frac{dE}{dx}>0$ in any direction from $P$ we would get the opposite conclusion, no matter which way we move the field, $E$, is increasing so at $P$ the field has a minimium value.

There are also some other cases, which are more complicated, but the upshot is that when you're at a point $P$ where $\frac{dE}{dx}=0$ you're at a point where the field is at is strongets or weakest (you should also look up global/local maximum/minimum).

KhanAcademy has a lot of helpful videos on this subject.

$\endgroup$

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.