1. (a)The solution to this problem invloves a simple equation from electrostatics for a point charge.
The equation is
E = kQ/r2
where
E = the electric field you want to find in NC-1 or Vm-1
(a) To solve the first part of the problem, you must have the charge of the iron nucleus. You must be aware that in magnitude the charge on a proton is the same as that of an electron but only the sign is opposite.
Further atomic number (Z) gives a representation for the charge of the nucleus and a neutral atom.
Here you're given
Z = 26
Therefore
Q = Ze
The electric field would be
E = (9 x 109)(4.16 x 10-18)/(6 x 10-10)2
(b)Using the same principle for the second part of the problem, you'll have
E = (9 x 109)(1.6 x 10-19)/(5.29 x 10-11)2
2. You have to get few basic facts clear before trying to understand my answer to your fascinating question.
The first and foremost principle is that electricity can only flow if there is a complete path for charges to flow. Normally air is a good insulator. However, there are occasions when this insulation is broken down such as with lightning that occurs naturally and arc lamps which are man made.
If charges are to flow then as we just said there must be a complete path or at least a network for the charges to pass on from one place to another. This is impossible in air unless you CREATE charges in the atmosphere by some articicial means that can conduct these and other charges. These charges can be created in the atmosphere when you by some means "TEAR" electrons off from the atoms in the atmosphere. This process is called IONIZATION and can created by various means such as supplying high electical energy as in the case of "arcing" or using radioactive sources such as radium and so on.
Here we are only interested in the former. Let's go into that in a little more detail. Normal air is neutral. By applying a high potential difference, you can ionize the atmosphere and at very high temperatures you have what's sometimes referred to as the fourth state of matter called PLASMA. Plasma is simply hot ionized gas. Charges of all kinds are floating around and it is belived that over 99% of the universe is made of plasma, our sun included!
Incidentally physicists have been trying in vain to harness this power of plasma (sun's source by nuclear fusion) by various techniques for over four decades. May be one day this source by "laser zapping" could be economically viable?
Coming back to the electical arc, have you understood the mechanism? You supply a high potential difference in between two junctions as in a carbon arc lamp, or the more common flurorescent lamp at homes and the ionization of the matter in between causes the familiar electrical arc.
Recall me telling you in the beginning that you have to understand a few principles? One more useful principle is how light is produced in general. Physicists usually explain this by turning to electron energy levels in an atom. Light is produced when an electron FALLS from a higher energy level to a lower energy level. The difference of energy between the two levels is what is emitted as light according to a famous physicist Max Planck who revolutionized physics with this suggestion first in 1900. Of course he did not have the courage to extend this and it was our dear old EINSTEIN who actually spread the concept in 1905.
The "arcing" is occuring because the ionized gas (plasma) has electons which are falling into lower energy levels.
I hope you've understood all this and do not hesitate to contact me should you have further questions on this.
When I was in school I think in Grade 7/8 my project was a simple carbon arc lamp. I was as fascinated with it then. I though you too may enjoy reading about the History of Electric Lighting Technology at the URL:
http://sheldonbrown.com/marty_light_hist.html
Pay special attention to the Carbon Arc Lamp which was my project and another useful link would be reading about the Discovery of the Carbon Arc, The Carbon Arc Experiment That Can Be Done in the Classroom (after taking the necessary precautions) and so on at the URL:
http://myron.sjsu.edu/caesars/CARBARC.HTM
I'm going to stop writing more now.
3. The answer to the first part of the problem involves the use of a very important equation in electostatics.
F = Q E
where
F = electric force in N
The other principle you must know is the DIRECTION of an electric field is traditionally taken to be the direction in which a UNIT POSITIVE CHARGE would move when placed at that point in the field.
a) You're given here F and Q
Making E the subject of the equation, you'll have
E = F/Q
= 6.20 x 10-9 /(55 x 10-6)
I've not taken the sign of the charge in the above equation to find out the magnitude of the force.
Obviously by looking at the principle above, if the charge is negative and experiences a downward force, the direction of the electric field here must be UPWARD. (negative charge attracted by the positive of the electric field, don't forget LIKE CHARGES REPEL and UNLIKE CHARGES ATTRACT).
b) Using the same equation above and the other I gave you last time Q = Ze, you'll have
F = (29)(1.6 x 10-19)(1.13 x 10-4)
ACTING UPWARD at that point in the electric field.
4. The first task for you will be drawing a straight line. Draw a rough sketch approximately to scale. On the left you should have q2 the unknown charge, in between the origin O and on the other side q1.
The formula to calculate the electric field is
E = kQ/r2
The electric field due to q1 would be
E = (9 x 109)(4 x 10-9)/(0.6)2
I would recommend that you solve for practice some of my other problems on electic field where I have elaborated on these rules.
(i) In the first part of the problem you want the net electric field to be 50 NC-1 in the +x-direction.
The formula is given by
Enet = E1 + E2
50 = 100 + E2
That is
E2 = - 50 N/C
= kq2/r2
Therefore
- 50 = (9 x 109)[q2]/(1.2)2
giving you
q2 = -8 nC
(ii) In the second part of the problem you want the net electric field to be 50 NC-1 in the -x-direction.
The formula once again is given by
Enet = E1 + E2
-50 = 100 + E2
That is
E2 = - 150 NC-1
= kq2/r2
Therefore
- 150 = (9 x 109)[q2]/(1.2)2
giving you
q2 = -24 nC
5. The first step you must undertake is to draw a rough sketch approximately to scale. If you did that, you'll find that the electic field due to the first charge is towards the positive x-axis. The field due to the second charge is along the third quadrant, making an angle of 53o below the negative x-axis.
You have to now resolve the field into the two perpendicular directions.
Considering only the second charge now. The electric field is
E2 = kq2/r2
where you've found the distance between the second charge and the origin using Pythagoras to be 2 m.
E2 = 4500 NC-1 53o below the -x axis
We can now resolve this into two components, one along the negative x-axis and the other along the negative y-axis, giving you
E2x = 4500cos53
and the negative y-direction
E2y = 3594 NC-1
The field due to the first charge as we said earlier will be along the positive x-axis and equal to
E1x = kq/r2
The resultant field along the x-direction is therefore
E = 6250 - 2708
E = 3594 NC-1 along the -y-direction.
b) To solve the second question you have to find the RESULTANT FORCE on a proton at the midpoint between q1 and q2.
This is done using the same principle and given by
Resultant F = F1 + F2
F1 will be vertically downward and have a magnitude given by Coulomb's law
F1 = kQq/r2
where you've taken the charge on the proton to be the standard value 1.6 x 10-19 C
and the force due to the second charge also will be vertially downward given by
F2 = kQq/r2
The total will be
Fnet = 6.75 x 10-15
Therefore using Newton's second law the acceleration
a = F/m
k = an electostatic constant = 9 x 109 N(mC)-2 in vacuum
Q = charge in C
= (26)(1.6 x 10-19)
= 4.16 x 10-18 C
= 1.04 x 1011 NC-1
= 5.15 x 1011 NC-1
Q = charge in C
E = electric field in NC-1
= 1.13 x 10-4 NC-1
= 5.2 x 10-22 N
= 100 NC-1 in the positive x direction.
= (9 x 109)(2 x 10-6)/(2)2
= 2708 NC-1
= (9 x 109)(1 x 10-6)/(1.2)2
= 6250 NC-1
= 3542 NC-1 along the +x-direction and
= (9 x 109)(1 x 10-6)(1.6 x 10-19)/(0.8)2
= 2.25 x 10-15 N
= 4.5 x 10-15 N
= (6.75 x 10-15)/(1.67 x 10-27)
= 4 x 1012 ms-2