Steven W. answered • 02/01/17
Physics Ph.D., college instructor (calc- and algebra-based)
Hi Porcia!
I am updating my answer based on your response below. This may be a little tricky with no diagrams, but perhaps you can refer to drawings in your textbook or online which may help with what I am describing.
To determine the electric field at any point from a charge distribution, you have to come up an expression for the electric field, dE, from an infinitesimal charge element, dq, of that distribution, and then integrate that expression over the distribution, using symmetry (as possible) to make the job easier.
Taking "the axis of the disk" (as we usually do) to mean the axis through the center of the disk perpendicular to its surface, the electric field from an infinitesimal charge element dq at a point P along the axis is:
dE = kdq/d^2
where d is the distance from the charge element to P. If the point P is a (perpendicular) distance z from the center of the disk, then we can write r above as
d = sqrt(r^2 + z^2) --> d^2 = (r^2 + z^2)
where r is the radius at which dq is on the disk.
Just for sake of argument, we can assume the charge on the disk is positive (nothing really changes if it is negative, except the resulting electric field points the other way). For a positive dq, its electric field dE and point P points AWAY from the infintesimal charge unit, at some angle PHI with respect to the optical axis. Here is the point that may be tricky to see without a diagram; it is where we can invoke a symmetry argument. For any infinitesimal charge element dq, there is a diametrically opposed element, also of magnitude dq, and at the same radius, which generates its own dE (equal in magnitude to the other dE) at point P. That dE points in such a way that its component perpendicular to the axis cancels out the perpendicular-to-the-axis component of the other dq. Because of this, the only component of any dE that survives the integration over all the dq's on the disk is the component parallel to the axis.
In terms of the angle PHI described above, the component of each dE parallel to the axis (dE_z) can be written:
dE_z = dE(cos(PHI)) = cos(PHI)(kdq/(r^2 + z^2))
Another tricky thing to see without a diagram may be that cos(PHI) may be written as
cos(PHI) = (z/sqrt(z^2 + r^2))
So dE_z becomes:
dE_z = z(kdq)/[(r^2 + z^2)^(3/2)]
Now, we need to integrate this over all the dq's. With this disk charge distribution, it is most convenient to do that with polar coordinates. But we still need to put dq in a form that depends on the polar coordinates. Since the disk is uniformly charged, we know that:
dq = (SIGMA)dA
where dA is an infinitesimal area unit and SIGMA is the charge per unit area (which is given). In polar coordinates, the infinitesimal area unit dA can be written as
dA = rdrd(THETA) (compare to dxdy, the area unit in Cartesian coordinates)
Putting all this into the dE_z expression, we get:
dE_z = kz(SIGMA)rdrd(THETA)/[(r^2 + z^2)^(3/2)]
which we can no integrate over the entire disk (in the r direction, from 0 to R, where R is the radius of the disk; and in THETA from 0 to 2pi, so go all the way around the disk).
Even though this may look daunting, we are helped by the fact that a lot of these quantities are constants. Also, since we integrate r and THETA independently, the THETA integral is just integrating d(THETA) from 0 to 2pi, which gives 2 pi. So integrating the above monstrosity reduces to integrating the expression:
E_z = (2pi)kz(SIGMA)*{integral from 0 to R of rdr/[(r^2 + z^2)^(3/2)]}
You can do the integral by u substitution (or other methods).
I will leave that there for now, so you can look it over and see if this makes sense, and then try the integral. Once you complete the integration, you should have an expression for the full electric field generated by the disk at point P, in terms of z. Then you can plug in the values, calculate E_z, and convert it ti MN/C.
If you have any other questions right away, please let me know. Also if you get stuck integrating, or would like to check an answer.
Steven W.
02/01/17
Porcia C.
02/01/17