Calculating induced charges by a charge between two capacitor plates.

[anirudhsharma1]

A charge Q is kept at a distance x from one of the plates of a parallel capacitor having plates d distance apart with cross section area A.The whole system is earthed.Find the value of induced charges on both the plates.

I tried putting the sum of induced charges equal to Q and balancing electrical fields but that led me nowhere!
help!

 

[FireStorm000]

Assuming the material between the plates is air or vacuum, the plates should charge such that the E field between them cancels the E field of the charge Q. Does that get you started? If you still need help, post what you do know, and what you've already tried. You have a problem statement, but not much else.

 

[anirudhsharma1]

i tried that too
but since the field due to an infinitely long charged surface doesn't depend on the distance
i don't see how this will be of help!

 

[Saitama]

i tried that too
but since the field due to an infinitely long charged surface doesn't depend on the distance
i don't see how this will be of help!

Ever heard about method of image charges?

 

[FireStorm000]

i tried that too
but since the field due to an infinitely long charged surface doesn't depend on the distance
i don't see how this will be of help!

Your problem statement says that the plates have finite area A. Now, if the capacitor is far enough away from the charge q, you should be able to pretend it's a single point; that is, if there are two plates, close to each-other, and far from q, you can pretend that the E field from q doesn't change over the area of the capacitor. Once you know that, you have a singe value for E field between the plates, and you just figure out how much charge would be on the plates in a capacitor of that configuration. Hope that helps.

-FireStorm-

 

[anirudhsharma1]

Ever heard about method of image charges?

nope
and did i forget to mention
the charge Q is placed BETWEEN the capacitor plates.

 

[anirudhsharma1]

Your problem statement says that the plates have finite area A. Now, if the capacitor is far enough away from the charge q, you should be able to pretend it's a single point; that is, if there are two plates, close to each-other, and far from q, you can pretend that the E field from q doesn't change over the area of the capacitor. Once you know that, you have a singe value for E field between the plates, and you just figure out how much charge would be on the plates in a capacitor of that configuration. Hope that helps.

-FireStorm-

Really sorry
but in my problem statement
i didnt mention that the charge Q is placed BETWEEN the two plates.

 

[FireStorm000]

Really sorry
but in my problem statement
i didnt mention that the charge Q is placed BETWEEN the two plates.

Hmmm, that may or may not make things easier, depending on the exact geometry of the problem. I don't suppose it's centered in the middle of the plates, equidistant from both?

 

[rude man]

Q = CV. If the whole system is earthed, what is V?

 

[anirudhsharma1]

Q = CV. If the whole system is earthed, what is V?

its not a question from capacitance.
from basic concepts of electrostatics.

 

[anirudhsharma1]

Hmmm, that may or may not make things easier, depending on the exact geometry of the problem. I don't suppose it's centered in the middle of the plates, equidistant from both?

nope
at a distance x from one of the plates.
i think that the method of image charges can be used to get the answer
i just don't know how to apply it
nor does my school teacher have any clue about it.

 

[rude man]

If both plates of the capacitor are grounded (earthed), what is the E field between the plates? So how much energy is required to move a charge Q form one plate to the other? Which tells you what about surface charges on the plates?

 

[FireStorm000]

nope
at a distance x from one of the plates.
i think that the method of image charges can be used to get the answer
i just don't know how to apply it
nor does my school teacher have any clue about it.

Calculus comes to mind here, and I'm not sure if you can solve this without it. You'll likely have to integrate infinitesimal charge over each capacitor plate in turn. As long as you know the location and strength of the point charge, you can solve for E field and therefore charge at locations along the plate. Let me know if you want help setting it up, or if you don't know calculus.

 

[FireStorm000]

If both plates of the capacitor are grounded (earthed), what is the E field between the plates? So how much energy is required to move a charge Q form one plate to the other? Which tells you what about surface charges on the plates?

I'd be surprised if that approach works when the point charge is between the plates, though I've been wrong before. In this case the plates aren't behaving like a capacitor, they are more behaving like independent grounded pieces of metal.

 

[rude man]

I'd be surprised if that approach works when the point charge is between the plates, though I've been wrong before. In this case the plates aren't behaving like a capacitor, they are more behaving like independent grounded pieces of metal.

If they're both grounded are they still independent? And two parallel plates, correctly described in your problem as a capacitor, do not a (shorted) capacitor make?

I too could be wrong but I know where I'd put my money ...

 

[FireStorm000]

If they're both grounded are they still independent? And two parallel plates, correctly described in your problem as a capacitor, do not a (shorted) capacitor make?

I too could be wrong but I know where I'd put my money ...

I don't think so; with the charge being placed in-between the plates, both plates should have the same polarity, either both positive or both negative, with ground absorbing the excess charge. The location of the point charge makes all the difference, because now the E field varies with position; there's no single E field between the plates.

 

[rude man]

OK., so assume both plates have a + charge on them facing each other. That means that their opposite faces have to have the same negative charge. But the opposite faces are effectively touching each other, so that's impossible: no charges allowed inside a perfect conductor!

 

[FireStorm000]

OK., so assume both plates have a + charge on them facing each other. That means that their opposite faces have to have the same negative charge. But the opposite faces are effectively touching each other, so that's impossible: no charges allowed inside a perfect conductor!

The fact that the plates are not only connected to each-other, but also to ground seems important to me. The E field should push charge out of both plates into ground. Not sure I can prove it without actually solving the problem though, and I was going to let the OP give it a try first.

 

[anirudhsharma1]

i think if the plates are grounded,
a negative charge must appear on the plates to make the total potential=0(grounded)
we can solve for both the plates separately(correct me if i am wrong).
should i give this a try??

 

[ehild]

I think you should try the method of images. When you have a single point charge in front of a grounded metal plate, the electric field between the point charge and the metal is the same as that of two point charges, one the real charge, the other is its image, at position which is the mirror image of the real charge and with equal but opposite charge. The difficulty is that you have two metal plates, so two image charges, and the image behind one plate is also mirrored by the other plate, so there is an infinite sequence of images behind both plates. 1 and 1' are the images of the real point charge, 2 is the image of 1' and 2' is the image of 1 and so on. The same you got if you stand between two parallel mirrors.
Of course, the electric field behind the grounded plates is zero.

ehild

 

[anirudhsharma1]

I think you should try the method of images. When you have a single point charge in front of a grounded metal plate, the electric field between the point charge and the metal is the same as that of two point charges, one the real charge, the other is its image, at position which is the mirror image of the real charge and with equal but opposite charge. The difficulty is that you have two metal plates, so two image charges, and the image behind one plate is also mirrored by the other plate, so there is an infinite sequence of images behind both plates. 1 and 1' are the images of the real point charge, 2 is the image of 1' and 2' is the image of 1 and so on. The same you got if you stand between two parallel mirrors.
Of course, the electric field behind the grounded plates is zero.

ehild

im confused.
first of all i don't know about the image charges method
and secondly the potential is zero too right??
which can correctly lead me to a solution.

 

[ehild]

The potential is zero on the plates, but it is not zero between the plates. ehild