2
$\begingroup$

Say you have a set of parallel plates, one is positive and one is negative, if you change the distance between them would electric field strength change or potential difference, given the equation

$E=dV/dx$

From pure intuition, I think electric field strength will change as the field line bulges out from the sides, and as the plates are far enough from each other, each plate can be treated as a point charge and the system becomes a dipole, and E decreases at 1/r^3:

picture: (sorry I didn't upload the picture here in case I violate copyright) http://www.electrobasic.com/uploads/3/2/3/4/32342637/9893979_orig.jpg

But I read from this post:Why does the potential difference between two charged plates increase as they move further apart? and it says E must remain constant, and I am not sure why that is, and I do not see how simply moving the plates apart will increase the potential difference, as

$V=kQ/r$

and as r increases, v should decrease, and I don't understand how simply moving them apart will store energy in the system.

So why must E remain constant and V increase as the plates are separated further?

Thanks.

$\endgroup$
2
$\begingroup$

If the plates have a fixed charge $±Q$ (equal and opposite) and they are isolated, then they will attract each other. When you move them apart, you are doing mechanical work (opposing the force of attraction).

As long as the plates are large compared to their separation, the field in between is roughly uniform, and it will remain so as they move further apart; this means that the force they experience will be the same as you increase the distance. This is not intuitive until you realize that most of the plate "might as well be infinite" as seen from a point charge on the opposite plate. It's only when the angle subtended by the plate becomes "noticeably less than" $2\pi$ that you start to see a drop off of force - as you said, the lines of the E field will start to bulge out. Another way to look at this is with a simple Gaussian pill box around the plate: the flux through the surface depends only on the charge in the box, which doesn't change.

$\endgroup$
1
$\begingroup$

If you treat parallel plates as infinite sheets, then you get a constant field in between that depends only on charge density. This isn't exactly true in reality, since real plates aren't infinite. But it gives a good estimation.

$\endgroup$
0
$\begingroup$

$C={kA\over d}$. When $d$ increases, $C$ decreases. $Q=CV$. $C$ stands for capacitance, and $V$ stands for potential difference. $C$ decreases, $V$ increases. That's why the potential difference increases.

$\endgroup$

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.