anirudhsharma1 said: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!
anirudhsharma1 said: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!
Pranav-Arora said:Ever heard about method of image charges?
FireStorm000 said: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-
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?anirudhsharma1 said:Really sorry
but in my problem statement
i didnt mention that the charge Q is placed BETWEEN the two plates.
rude man said:Q = CV. If the whole system is earthed, what is V?
FireStorm000 said: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?
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.anirudhsharma1 said: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 said: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 said: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 said: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 ...
rude man said: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!
ehild said: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
The formula for calculating induced charges between two capacitor plates is Q = CV, where Q is the charge, C is the capacitance, and V is the voltage.
The polarity of the induced charges is determined by the direction of the electric field between the two capacitor plates. If the electric field is directed from positive to negative, the induced charges will be positive on one plate and negative on the other.
The magnitude of induced charges is affected by the distance between the two plates, the voltage applied, and the dielectric constant of the material between the plates. A larger distance, higher voltage, or higher dielectric constant will result in a larger magnitude of induced charges.
Yes, induced charges can be manipulated by changing the voltage applied or the distance between the plates. This is the principle behind variable capacitors, which are used to control the amount of charge stored between two plates.
When the voltage is removed, the induced charges will redistribute themselves evenly between the two plates, resulting in a neutral charge on each plate. This is known as discharging the capacitor.