Complete Question
A thin, horizontal, 12-cm-diameter copper plate is charged to 4.4 nC . Assume that the electrons are uniformly distributed on the surface. What is the strength of the electric field 0.1 mm above the center of the top surface of the plate?
Answer:
The values is 
Explanation:
From the question we are told that
The diameter is 
The charge is 
The distance from the center is 
Generally the radius is mathematically represented as

=> 
=> 
Generally electric field is mathematically represented as
![E = \frac{Q}{ 2\epsilon_o } [1 - \frac{k}{\sqrt{r^2 + k^2 } } ]](https://tex.z-dn.net/?f=E%20%3D%20%20%5Cfrac%7BQ%7D%7B%202%5Cepsilon_o%20%7D%20%5B1%20-%20%5Cfrac%7Bk%7D%7B%5Csqrt%7Br%5E2%20%2B%20%20k%5E2%20%7D%20%7D%20%5D)
substituting values
![E = \frac{4.4 *10^{-9}}{ 2* (8.85*10^{-12}) } [1 - \frac{(1.00 *10^{-4})}{\sqrt{(0.06)^2 + (1.0*10^{-4})^2 } } ]](https://tex.z-dn.net/?f=E%20%3D%20%20%5Cfrac%7B4.4%20%2A10%5E%7B-9%7D%7D%7B%202%2A%20%288.85%2A10%5E%7B-12%7D%29%20%7D%20%5B1%20-%20%5Cfrac%7B%281.00%20%2A10%5E%7B-4%7D%29%7D%7B%5Csqrt%7B%280.06%29%5E2%20%2B%20%20%281.0%2A10%5E%7B-4%7D%29%5E2%20%7D%20%7D%20%5D)

Answer:
System D --> System C --> System A --> System B
Explanation:
The gravitational force between two masses m1, m2 separated by a distance r is given by:

where G is the gravitational constant. Let's apply this formula to each case now to calculate the relative force for each system:
System A has masses m and m separated by a distance r:

system B has masses m and 2m separated by a distance 2r:

system C has masses 2m and 3m separated by a distance 2r:

system D has masses 4m and 5m separated by a distance 3r:

Now, by looking at the 4 different forces, we can rank them from the greatest to the smallest force, and we find:
System D --> System C --> System A --> System B
No, that's silly.
You've got your Pfund series where electrons fall down to the 5th level,
your Brackett series where they fall to the 4th level, and your Paschen
series where they fall to the 3rd level. All of those transitions ploop out
photons at Infrared wavelengths.
THEN next you get your Balmer series, where the electrons fall in
to the 2nd level. Most of those are at visible wavelengths, but even
a few of the Balmer transitions are in the Ultraviolet.
And then there's the Lyman series, where electrons fall all the way
down to the #1 level. Those are ALL in the ultraviolet.