In the figure below, an electron (e) is to be released from rest on the central
ID: 2272975 • Letter: I
Question
In the figure below, an electron (e) is to be released from rest on the central axis of a uniformly charged disk of radius R. The surface charge density on the disk is +1.20
In the figure below, an electron (e) is to be released from rest on the central axis of a uniformly charged disk of radius R. The surface charge density on the disk is +1.20 mu C/m2. What is the magnitude of the electron's initial acceleration if it is released at a distance R from the center of the disk? What is the magnitude if it is released at a distance R/100 from the center? What is the magnitude if it is released at a distance R/1000 from the center? Why does the acceleration magnitude increase only slightly as the release point is moved closer to the disk?Explanation / Answer
The electric potencial at the axis can be calculated by means of an integral.
V(D) = sigma / (4 pi eps0) integration(phi from 0 to 2 pi) dphi integration(r from 0 to R) r dr / sqrt(r^2+D^2) =
= [sigma /(2 eps0)] [sqrt(R^2+D^2) - D]
Then, the electric field is Ez(D) = -dV/dD =
= [sigma /(2 eps0)] [ 1- D / sqrt(R^2+D^2)]
If D=R/N
Ez(R/N) = [sigma /(2 eps0)] [ 1- 1 / sqrt(1+N^2)] ...................................................(1)
The force on an electron would be
F = e Ez, where e is the electron charge,
and the electron acceleration
a = e Ez / m,......................................(2)
where m is the electron mass.
(a) N=1, put this value in (1) than for "a" put in (2)
(b) N=100, similiar to (a)
(c) N=1000, similiar to (a)
(d) as oon as the electron approch the central point on disk....the electric field continously attract and electric field increse hense as a result its increase slightly for each step