Electric field due to a charged disc #4
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Electric field due to a
Electric field due to a charged disc sets up an electric field around it. You need to imagine the disc as a collection of rings to derive the formula.
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Well, if this disc has a charged surface, it is bound to set up an electric field around it…so let us go ahead and try to find the electric field at a point P on the z axis at a distance z from the center of the disk. Now what you would see here is that the situation is quite comparable to finding electric field due to a charged ring…all you need to do is imagine that this disc is a collection of rings and you have to sum up the electric field due to each ring at point P. And of course after doing the earlier lesson, you know how to find charge at point P, due to a each of these rings.
So let us take a ring on the disk at some radius r and we will assume the ring is so thin that we can treat the charge on the ring as very small and equal to dq… now to establish the electric field dE generated by this ring at point P we can use the equation established earlier and say that dE due to this ring is
And you also know from the earlier lesson that the direction of the electric field due to this ring will point in upward direction. Now to find the total E at this point, we need to sum up contribution due to each ring… well we can see that the radius of various rings on this disc change from a radius of zero at the center to a radius of R for the outermost ring…and by now you’d have a sense that we will need to make use of integration with r value changing from zero to R.
So if radius is the variable we have to deal with, we need to pull in dR in this equation to enable us to apply integration to r. So assuming the thickness of the ring is very less, lets say it is dr, we could say the surface area of such a ring is its circumference X the thickness dr. or dA = 2pi r dr
And since we know that row is charge per unit area, row = dq/dA or dq = row dA, we can then say that this equals row 2 pi r dr.
And if we plug in this value of dq in this equation, what we get is
Now if you do the integration of this expression, what you get is
But for those of you who are a little more mathematically inclined I’d do a quick derivation of this integration for you.
And when you apply limits, what you get is
So this is the value of the electric field at a distance z produced by a flat disk at point P. So let us play around with this formula and see what would be the value of electric field at P if we had an infinitely large disc here. Well an infinitely larger disc would mean R tends to infinty in which case, this expression would become zero. And what you get is
And this expression for field produced by such a disc is same as what we saw in the earlier lesson for a field produced by an infinitely large plate…becaues an infinitley large plate would look the same as an infinitely large disc to a point charge at P
We could also keep R finite, the way it is here, and make z infinitely large and you would see is that again this expression vanishes and you get the same answer E = sigma by 2 epsilon
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Well, if this disc has a charged surface, it is bound to set up an electric field around it…so let us go ahead and try to find the electric field at a point P on the z axis at a distance z from the center of the disk. Now what you would see here is that the situation is quite comparable to finding electric field due to a charged ring…all you need to do is imagine that this disc is a collection of rings and you have to sum up the electric field due to each ring at point P. And of course after doing the earlier lesson, you know how to find charge at point P, due to a each of these rings.
So let us take a ring on the disk at some radius r and we will assume the ring is so thin that we can treat the charge on the ring as very small and equal to dq… now to establish the electric field dE generated by this ring at point P we can use the equation established earlier and say that dE due to this ring is
And you also know from the earlier lesson that the direction of the electric field due to this ring will point in upward direction. Now to find the total E at this point, we need to sum up contribution due to each ring… well we can see that the radius of various rings on this disc change from a radius of zero at the center to a radius of R for the outermost ring…and by now you’d have a sense that we will need to make use of integration with r value changing from zero to R.
So if radius is the variable we have to deal with, we need to pull in dR in this equation to enable us to apply integration to r. So assuming the thickness of the ring is very less, lets say it is dr, we could say the surface area of such a ring is its circumference X the thickness dr. or dA = 2pi r dr
And since we know that row is charge per unit area, row = dq/dA or dq = row dA, we can then say that this equals row 2 pi r dr.
And if we plug in this value of dq in this equation, what we get is
Now if you do the integration of this expression, what you get is
But for those of you who are a little more mathematically inclined I’d do a quick derivation of this integration for you.
And when you apply limits, what you get is
So this is the value of the electric field at a distance z produced by a flat disk at point P. So let us play around with this formula and see what would be the value of electric field at P if we had an infinitely large disc here. Well an infinitely larger disc would mean R tends to infinty in which case, this expression would become zero. And what you get is
And this expression for field produced by such a disc is same as what we saw in the earlier lesson for a field produced by an infinitely large plate…becaues an infinitley large plate would look the same as an infinitely large disc to a point charge at P
We could also keep R finite, the way it is here, and make z infinitely large and you would see is that again this expression vanishes and you get the same answer E = sigma by 2 epsilon
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