Electrostatics Examples

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Prepared by

ARCHITECT RAISON JOHN J. BASSIG

1^{st Placer ? June 2006 PRC Architect Licensure Examination}

^{10th Placer ? March 2007 PRC Master Plumber Licensure Examination}

^{So you are about to take the board exam. Nervous? You should be.}

^{A few months before I took mine, I have experienced what every board examinee goes through: Unexplainable nervousness. Butteflies in the stomach. Catatonic looks.}

^{Am I ready?}

^{For one thing, it's better to be nervous NOW than to feel your sweat drips and your veins popping ON the day of the exam itself.}

^{Imagine yourself staring blankly at those test questionnaires. 45 minutes have passed. You are already at question number 31, yet your answer sheet remains blank - a complete mental blackout.}

^{Your mind then wanders. As you look along the walls of the quiet classroom, you begin to reminisce your days in college. How you wish you remembered what that professor had taught you. You gaze up the ceiling and wonder about the future of your career, hanging in the balance, through these test questions you are holding right now. You inhale deeply and pour out a sigh of air. You pray for luck, hoping the next question would be something that you have read in your 1-week review. You turn the next page.}

^{Tsk. Where in God's name did they get all these questions???}

^{Mi... ni... mi...ni... mai... ni... mu.}

^{This is a typical story of what could go through a mind of a board examinee. A board examinee who did not prepare, relied on luck, and had no idea how to go about answering the board exam questions. Thankfully, that did not happen to me. When I took the Board Exams for Architects in June 2006 and for Master Plumbers in March 2007, my confidence was sky-high. All because I was prepared. This leads me to my conclusion: "Talo ng preparado ang matalino."}

^{So for almost 12 months now, I have been sharing to others through seminars, lectures and informal discussions, how I prepared for the board examination. Allow me to share with you some tips and tricks based on my own experience that can help you prepare for your own review.}

^{SETTING YOUR GOAL}

^{The first and most important tip that I would give you is SET YOUR GOAL. Whether you just want to pass and get a license, or maybe land a place among the top 10, or probably get a grade of 90+ on the most dreaded subject, it helps if you have direction. The more specific your goal is the better. Try to write it on paper and feel committed to it. Mine was to top the board exams - and I did.}

^{Now in order for you to work on that goal, you must have the RIGHT MOTIVATION. This is the second and equally important tip. Your motivation is your drive in achieving your goal. It could be anything from a person that inspires you (an idol, a crush, or a loved one) to a material promise (like an all-expense out-of-the-country trip or a brand new car). It's up to you! Use this to keep you going everyday. Use this to get you up early in the morning when you normally wake up late. When you are motivated, you feel like you can do anything. Yes, ANYTHING! So, this gives you a psychological advantage.}

^{The board examination, just like any other sport or competition, is all about how you prepare for a fight. It is all about setting your mind, body, and emotions straight before the main event. During the preparations, YOU ARE YOUR OWN ENEMY. The less you know about yourself, the harder it is to succeed in this undertaking. That is why you have to train hard. What you do to your time is all up to you. You could sit back and relax if you want, but keep in mind, all the others might have already gone far ahead in their reviews while you are still sitting pretty waiting for that right moment.}

^{DECIDING WHEN TO TAKE THE EXAM}

^{Time is one of the most critical resources in your entire preparation. Once time goes by, it can never be brought back. So speaking of time, the first question you have to ask yourself is: "WHEN AM I GOING TO TAKE THE EXAM?" It may sound simple but believe me, this simple question will determine how much extraordinary effort you may have to exert throughout the rest of your review.}

^{I know some people who spent 4 months studying, paid a lot of money for their review classes, only to back out 3 weeks before the exam. For some unexplainable reason, they found themselves "not yet ready" and would just opt to take the next board exam, which is scheduled 6 months or 1 year later. Uh-oh. Now that's a lot of time gap until the next exam. You would easily forget most of the topics discussed in the review center and you would be forced to take refresher courses again later on. If they were unsure, they should have not enrolled immediately. It's a waste of time and a waste of money. In my opinion, the only logical explanation for their backing out is that they were half-hearted with their review. You know, once you have decided that you are going to take the board -- GO FULL-FORCE!! As the saying goes, "Todo Na 'To!". Never ever be half-hearted. That's why the sooner you decide, the better your chances of making the most out of your time. In my case, I decided to take the June 2006 Board Exam for Architects as early as November 2005. I had 8 months preparation.}

^{KNOW THE COVERAGE OF THE EXAM}

^{YOU SHOULD KNOW what subjects you are to tackle in your review. But going back to your review, you should determine which subjects covered by the exam are you most comfortable with and which subjects are you having problems with. It's a good strategy to study your favorite subjects first. These subjects are usually the ones you are really interested in. The idea is to get you "IN THE ZONE". Think of yourself as a car that needs a jump start. Once you get going on one subject, then that's the time you absorb the other subjects along the way.}

^{KNOW YOURSELF}

^{The key to a successful board exam preparation is knowing one's self. Find out your strengths and your weaknesses. Are you a fast reader? Do you easily comprehend anything that you read? How good are you at numbers? Can you memorize dates well? Would you appreciate a book by looking at illustrations or by reading the article itself?}

^{You have to know yourself and seek for self-improvement. There are a lot of ways for you to improve. Use your strengths to concentrate on building up your weak points.}

^{TOTAL CONDITIONING}

^{Believe it or not, preparing for the board is not just a walk in the park (especially if you have set high goals in the beginning). If you really want to make the most out of your review and achieve the goals you have set, then you must be prepared in all aspects, and I really mean ALL: physical, emotional, spiritual, social, mental and even financial :)}

^{Most of us are night-owls, we begin to study at night until the wee hours of the morning. This is our comfort zone. This is where we feel we are more progressive studying. That's ok as long as you still have 6 to 8 hours of sleep. I usually study from}^11pm^{up to}^{6 am}^{the next morning, then sleep and wake up around}^{2 pm.}^{Whatever your comfy zone is, feel free to use it as you please. But as the board exam nears, there is a need to re-adjust your study habits. About 2 to 3 weeks before the exam, you have to practice waking up early and studying in the morning - not at night anymore. The idea here is to get yourself comfortable "thinking" in the morning because the board exam starts in the morning. Else, you'll feel sleepy and nauseaous during the exam. Also, be conscious of the food you eat 1 week before the exam. Nobody wants to have an upset stomach on the day of the exam.}

^{The hard-core reviewing should end at least 2 weeks before the exam. By this time, you should have studied what you needed to study. The 2 remaining weeks should give you time to relax and psych yourself. Ideally, a week before the exam, you should just be browsing through your notes to refresh your memory. No more all-nighters. No more burning your eyebrows. So don't stay up all night exhausting your energy cramming for more information. That was my mistake when I took the Board Exam for Master Plumber last March. Since I only had a month's preparation (because I was also working), I had to exert more effort in my studies - more than what my body can take. Come board exams, I was on Tylenol because of a fever. Sleep early the night before the exam.}

^{A board examinee must also expect pressure from a lot of people. There are pressures from your family, there are pressures from your relatives (especially those successful ones), there are pressures from your friends and barkadas, from your office and even from your girlfriend/boyfriend. The moment you decide you'll take the board exam and tell it to someone, word really travels fast. They would be greeting you 'good luck', but others would also expect you to do really good. I know some people who did not tell a single soul that they'll take the board exam. Maybe they were too pressured or shy. You don't have to be mum about it. Besides, whether you pass or fail, in time, everybody will still know that you took the exams. TAKE ALL THESE PRESSURES AND PUT IT TO YOUR ADVANTAGE. If you are motivated, these pressures would even help propel you to study harder.}

^{Depends on your religious belief, it's already a habit that board examinees be going to Baclaran or celebrate Holy Week in Manaog or complete the}^midnight^{mass during advent. We would even buy paraphernalia and bring our pencils, rulers, and books blessed. Prayers help a lot. Ask for His divine guidance. And pray for others too. But always remember all these will be useless if not complemented by studying. As the saying goes: 'genius is 1% inspiration, and 99% perspiration.'}

^{TRIPLE-CHECK TWICE AND ORGANIZE}

^{This might still be a bit of an understatement, but check, check, check, check your review materials again and again and again. Be wary of misinformation (errors on handouts, outdated information, etc.). It happens. Nothing is more sour than spending a lot of time memorizing a table of information from a code or law only to find out that it is already amended by a newer one. Equally damaging is reviewing notes you photocopied from your classmate only to find out that it was a typographical error on his part. The only way to be safe from this is to triple-check your review materials. Counter-check every information from other sources. DO NOT RELY ON ONE REFERENCE ALONE. Try to check other books and cross-reference it. Search the internet. Have discussions with some professors or friends about inconsistent info.}

^{During the course of your review, it is true that you have to make the most out of it and grab as much information as you possibly can. But it is also true that you can only take as much information for a certain period of time. This happens a lot. We borrow and photocopy a lot of books and handouts only to find itself stacked up on your bedroom floor.}

^{The index card would be your best friend in your review. Put information that you need to memorize in index cards and always keep them handy. You could browse through them while riding the bus or jeep. Use color-coded pens for highlighting old and new data. Use post-it on books to help you keep track of notes and points of discussions.}

^{PRIORITIZING THINGS}

^{An effective board exam review relies on discipline on the examinees part. There are so many temptations around: your favorite TV shows, the billiards hall, online gaming, etc. Learn how to prioritize. Think of the board exam as a one-time deal (although if you fail, you can take the board exam again - but nobody really wants to fail, right?) YOU WILL HAVE ALL THE TIME IN THE WORLD AFTER THE BOARD EXAM. Use your time wisely. Instead of watching TV for 3 hours, you could have understood and memorized a large portion of the book your reading already.}

^{MEMORIZATION TECHNIQUES}

^{If you have good photographic memory, you have an advantage. Equally advantageous is if you are a fast reader with good reading comprehension.}

^{Try using play of words and numbers through word association. Example of questions in our review: "Who was is the architect of}^Greenbelt^{2?", "Who is the architect of}^Greenbelt^{3?". Through word association, I know that the architect of}^Greenbelt^{2 is "Recio + Casas" because they are two (2) individuals (note the number 2 in}^Greenbelt^{). For}^Greenbelt^{3, I know the architect is G.F. Formoso because I associate the 3G phone (note the G in G.F. Formoso).}

^{Use letters as symbols}^{. Here's another example that appeared in the Master Plumber exam. "Which is denoted by its Inside Diameter? a) Pipe or b) Tube"}

^{I know that the PIPE is denoted by its Inside Diameter because I always write it "pIpe" (note the capital I, meaning Inside). I know the the TUBE is denoted by its Outside Diameter because I always write it "tubO" (tubo is tagalog for tube, note the capital O, meaning Outside). Get the idea?}

^{If your having trouble with numbers (like dates and laws) but good in memorizing visual patterns, try using the cellphone keypad as a guide. You'll remember the pattern you create while typing in the numbers more than the date or law itself.}

^{Rearrange information found in books.}^{Most of the time what is printed in books is not the best way how you would read and memorize it. Take the case of the National Building Code. I was memorizing 30+ types of pipe and their colors. In the book, it is arranged alphabetically by pipe type. What I did was I arranged it by pipe color. I found out that there are only 4 pipe colors so its easier to memorize. So I got 4 pieces of paper, painted them Red,}^Orange^{, Yellow, and Green respectively, and wrote in each paper the corresponding pipes. Voila. Easier to remember and understand.}

^{There are lots of other tricks like forming Acronyms or Creating a Song or Story. Try to create your own memorization tricks up your sleeves.}

^{BOARD EXAM: IS IT JUST MEMORIZING?}

^{Most people would say that the board examination is not really a gauge on how good a person is. They say it only reflects how good a person can memorize. Well, they are partly correct.}

^{But then again, memorizing is not as easy as A-B-C. It takes hardwork and a lot of patience. The most common mistake of board examinees is that they memorize just for the sake of 'enumerating' it. If you ask them what it's all about or how it works, they could not fully explain. The basic trick is: DO NOT MEMORIZE; BUT FAMILIARIZE. When you are familiar with the topic, you try to understand. When you fully understand, you try to analyze. You ask questions, "Why is this so?" and "Why is that so?". Then after you have achieved all this, you'll find it easier to memorize.}

^{The board exam is about setting a goal and working hard to achieve it.}

^{It's not just about memorizing. Nor is it about mi... ni... mi...ni... mai... ni... mu.}

^{So are you ready?}

^{ALL THE BEST}

TRANSITION METALS INCLUDING LANTHANIDES

The elements lying between s- and p-block elements of the periodic table are collectively known as transition or transitional elements(T.E?s).These elements either in their atomic state or in any of their common oxidation state have partly filled(n-1) d orbitals of (n-1)th main shell. In these elements the differentiating electron enters(n-1)dorbitals of (n-1)d orbitals of (n-1) th main shell and as such these are called d-block elements. This diefinition covers coinage metals, Cu,Ag and Au as transition metals because in their commonly occuring oxidation states, they have partly filled d-sub shells. However, it may be noted that the above definition does not include elements of group 12 i.e., Zn,cd and Hg.

Electronic configurations: All d- block elements show(n-1_d1-10 ns1-2 electronic, configuration. The half filled and completely filled d-orbitals acquire extra-stability and are favoured whenever possible.

General Characteristics of transition elements

1. Metallic Character.All the transition elements are metals and have high density, hardness, high tensile strength, ductility,malleability, high thermal and electrical conductivities and lustre. The metallic character is due to low ionization energies and a number of vacant orbitals in their atoms. The transition metals have very high melting points and boliling points. The high melting points of the transition metals suggest that the atoms in these elements are held together by strog metallic bonds which are formed due to the interaction of electorns in the outermost orbitals. The strength of bonding is roughly related to the number of unpaired electrons, stronger is the metallic bonding and consequently melting points are high. For example in transition series chromium has high melting point because it has five unpaired electrons.

2. Atomic and ionic radii: The atomic radii of elements of a particular transition series decrease with increase in atomic number but this decrease in atomic radius becomes small after midway. The decrease in size is attributed to increase in nuclear charge. However the effect of increased nuclear charge is partly cancelled by the increased screening effect of the electrons in the d-orbitals of penultimate shell. In the first transition series, the atomic radii become almost constant after chromium because the two factors, i.e., increase in nuclear charge and increase in screening effect just balance each other. The ionic radii also follow the similar trend.

3. Ionization Energies: The ionization energies of these elements are high, and in most cases lie between those of s-and p-block elements. This indicates that the transition elements are less electropositive than s-block elements. The ionization energy in any transition series effect of increased nuclear charge is partly neutralized by the extra-screening of the valence electrons from the nuclear attraction by the electrons being progressively added in the inner orbitals.

4. Oxidation States: Most of the transition elements show several oxidation states i.e.,they show variable valency in their compounds.

The outermost electronic configuration of the transition elements is(n-1)d^1-10ns². Since, the energy levels of (n-1)d and ns-orbitals are quite close to each other, hence both the ns-and (n-1) d-electrons are available for bonding purposes. Hence, the number of oxidation states displayed by these elements depends upon the number of d-electrons it has. For example, Sc having a configuration 3d14s2 may show an oxidation state of +2(only s- electrons are lost) and +3 (when d- electrons is also lost). The highest oxidation state which element of this group might show is given by the total number of ns- and unpaired(n-1)d-electrons.

5. Coloured Ions: Most of the transition metal compounds are coloured both in the solid state and in aqueous solutions. The colour of the compounds of transition metals may be attributed to the presence of incomplete(n-)d-subshell. The transition metal ions containing unpaired electrons can be easily promoted from one energy level to another in the same d-subshell. The amount of energy required to excite some of the electrons to higher energy states within the same d-subshell corresponds to energy of certain colours of visible light. Therefore, when light fails on a transition metal compounds, some of its energy corresponding to a certain colour, is absorbed causing excitation of d-electrons. The excess of other colours constituting white light are transimited and the compound appears coloured.

The transition metal ions which have completely filled d-orbitals are colourless. There are no vacant d-orbitals to permit excitation of electrons. Thus; Zn²⁺(3d¹⁰),Cd²⁺(4d¹⁰) and Hg²⁺(5d¹⁰) ions are colourless. The transition metal ions which have completely empty d-orbitals are also colourless. Thus Sc³⁺ and Ti⁴⁺ ions are colourless.

6. Magnetic Properties: The substances, which contain some species(atoms, ions or molecules) with unpaired electrons in their orbitals, behave as paramagnetic substances. Such substances are attracted by magnetic field. On the other hand, the substances whose constituent particles do not contain any unpaired electrons are repelled by magnetic field and are called diamagnetic. The transition metal ions generally contain one or more unpaired electrons in themand hence their compounds are generally paramagentic. The magnetic character is expressed in terms of magnetic moment. Larger the number of unpaired electrons in a substance greater is the paramagentic character and larger is the magentic moment. The magnetic moment is expressed by Bohr magneton and abbreviated as B.M.

7. Tendency to form complexes: Transition metals have a great tendency to form complexes because the size of their ions is very samll and for this particular reason they have a very high charged density. Moreover, they have vacant orbitals of right type of energy to accept lone pairs of electrons.

GENERAL PROPERTIES OF FIRST ROW TRANSITION METAL COMPOUNDS

1.Oxides: First row transition metals react with oxygen at elevated temperatures to form oxides of the type M₂O,MO,M₂O₃,MO₂,M₂O₅,MO₃ and M₂O₇. When the metal is in the low oxidation state, the oxides are usually basic. At intermediate oxidation states and high oxidation states, the oxides are normally amphoteric and acidic respectively. Some of these oxides dissolve in acid and alkalies to form oxometallic salts like potassium dichromate and potassium permanganate etc.

Potassium dichromate, K₂Cr₂O₇

It is prepared from chromite ore (FeCr₂O₇)

4FeCr₂O₄ + 16NaOH + 7O₂ 8Na₂CrO₄ + 2Fe₂O₃ + 8H₂O

2Na₂CrO₄ + H₂SO₄ Na₂Cr₂O₇ +Na₂SO₄ + H₂O

(orange)

Na₂Cr₂O₇+2KCl K₂Cr₂O₇ + 2NaCl

Potassium dichromate is an orange-red crystalline substance(m.p. 671 K). An aqueous solution of dichromate on treating with alkali turns yellow due to the formation of CrO₄^2- ions which on acidifying gives again Cr₂O₇^2- ions.

Cr₂O₇^2- + 2OH^- 2CrO₄^2- + H₂O

2CrO₄^2- + 2H⁺ Cr₂O₇^2-+ H₂O

It acts as strong oxidizing agent in acidic mediu

K₂Cr₂O₇ + 4H₂SO₄ K₂SO₄ + Cr₂(SO₄)₃ + 4H₂O + 3[O]

Cr₂O₇^2- + 14 H⁺ + 6e^- 2Cr³⁺ + 7H₂O[E^o = 1.31 V]

When any chloride is heated with K₂Cr₂O₇and Conc.H₂SO₄, orange-red fumes of chromyl chloride(CrO₂O₇) are formed

K₂Cr₂O₇+4NaCl + 6H₂SO₄ 2KHSO₄ + 4NaHSO₄+2CrO₂Cl₂ +3H₂O

Potassium Permanganate, KMnO₄

It is prepared from pyrolusite(MnO₂):

heat

2MnO₂ + 4KOH + O₂ 2K₂MNO₄ + 2H₂O

2K₂MnO₄ + Cl₂ 2KMnO₄ + 2KCl

2K₂MnO₄ + O₃ +H₂O 2KMnO₄ + 2KOH + O₂

Potassium permanganate is a purple coloured crystalline solid having melting point 523 K. It is slighlty soluble in water. It is a powerful oxidising agent in acidic, alkaline and neutral media.

Halides: First ?row transition elements react with halogens at high temperatures to form halides. The reactivity of halogens obeys the order F>Cl>Br>I. Usually fluorides are ionic. Bromides, chlorides and iodides become excessively covalnet with the atomic mass of the halogens.

Sulphides: Sulphides of transition metals are prepared either by direct combination of sulphur with metals or by the reaction of H₂S or Na₂S with a solution of their salts. In the sulphides oxidation state of metal is generally low because sulphur is a weak oxidising agent.

GENRAL PROPERTIES OF SECOND AND THIRD ROW TRANSITION ELEMENTS

Group 3 metals Y and La do not show transitional characterisitcs. In Group 4 Zr and Hf and their alloys are used in nuclear power enegineering. In group 5 alloys of Nb and Ta are used in jet enegines. IN group 6 both Mo and W with carbon, nitrogen, boron and silicon give hard refractory and chemically inert interstital compounds. In manganise group Tc and Re are very rare metals. Group 8,9,10 metals constitute three traids out of which first traids are known as ferrous metals.While elements of second and third traids arer known as platinum metals. In group 11 silver and gold are coinage metals used in making jewellery, coins and in dental filling. Group 12 metal Cd and Hg are used in making alloys and amalgams.

INNER TRANSITION ELEMENTS

The elements in which the last electron eneters in (n-2) f-orbitals are called f-block elements or inner transition elements. There are two series of f- block elements.

(i) Lanthanides: These are the elements in which differentiating electrons enter one of the 4f ?orbitals.

(ii) Actnides: These are the elements in which differentiating electrons enter one of the 5f-orbitals.

we observe that there is a general decrease in the ionic radii in the lanthanide series. The regular decrease in the size of lanthanide ions is known as Lanthanide Contraction.

Statements at a glance

1. Gravitational force is weaker but dominating than Coulomb force.

2. When two identical spheres carrying unequal +ve charges are brought in contact, charges will be equally distributed and force of repulsion will increase.

3. A ?vely charged rod can attract an uncharged ball and +ve ball.

4. Work done in taking a charge along a closed path is zero.

5. Field lines cannot be inclined to surface.

6. Field lines cannot form closed loops

7. Field lines should be perpendicular to surface.

8. Field due to a point charge varies inversely to r2

9. Field due to dipole varies inversely to r3

10. Electric field is parallel to dipole along axial line and along equatorial line.

11. Electric field is twice along axial line than in equatorial line.

12. Net force is zero on an electric dipole placed in uniform electric field.

13. At the centre of a dipole there is field, but potential is zero.

14. The value of potential at infinite distance from +ve or-ve charge is zero.

15. In midway b/w two equal like charges, electric field is zero, but potential is non zero.

16. Work done in equipotential surface is zero.

17. Stable equilibrium, on dipole in uniform electric field whn torque is zero, ?=0, dipole moment is in the direction of EF.

18. Unstable equilibrium, whn torque is zero, but, ?=180, dipole moment is opposite to EF.

19. Rotating a dipole once completely, wd is zero.

20. From stable equilibrium to unstable equilibrium wd is 2PE.

21. WD is zero

A] Whn a charge moves in a circle with another charge kept at the centre.

B] Whn a charge moves along an ellipse kept a charge at centre.

C] Whn a dipole is rotated once completely in another dipole field.

22. The flux depends only on net charge enclosed by the closed surface, not on position of charges

23. EMF is the difference in potential b/w the terminals of the cells whn circuit is open, potential difference is the difference in potential b/w any points of a circuit whn the circuit is closed.

24. In metals current carriers are electrons.

25. In electrolytes nd in gases current carriers are +vely nd ? vely charged ions.

26. In semi conductors current carriers are +vely nd ?vely charged holes.

27. The current due to random motion of free electrons is zero.

28. Electric current established with speed of light so every point of circuit experiences an electron drift. So current produces instantly.

29. The slope of VI graph gives R.

30. Alloys hav greater resistance than metals.

31. Alloys hav high resistivity nd zero temp. coefficient

32. Semiconductors hav ?ve temp. coefficient

33. Whn a battery is short circuited, then the entire power delivered by the battery is dissipated inside it as heat energy.

34. Whn internal resistance is small, maximum current is drawn from a battery.

35. Whn temp. increases internal resistance decreases.

36. In a battery, high internal resistance is used for safety.

37. Rheostat is used to obtain maximum balancing length by adjusting it.

38. Whn galvanometer shows 0 deflections, no current flow thrgh potentiometer.

39. Increasing length of potentiometer wire, increases its sensitivity bcoz potential gradient is small.

40. Resistance of copper is small, so it is used for potentiometer wire. If it is short circuited no pd will be available.

41. In series 60w glows brighter than 100w. becoz P ? R.

42. In parallel 100w bulb glows brighter than 60w. becoz V kept constant ? 1/R

43. Magnetic field is zero at any point along a straight conductor.

44. +ve charge cannot move along magnetic line of force, since direction of force is perpendicular to both velocity and magnetic field.

45. Lorentz force is perpendicular to velocity so on tangential acceleration and KE remain constant

46. Wd by lorentz force is zero becoz force nd displacement are perpendicular to each other.

47. Path of charged particle:

Parallel to mag.f ? straight line path.

Perpendicular to mag.f ? circular path.

At an angle wid mag.f ? helical.

48.Wd in full rotation for a mag. Dipole in a uniform mag. Field is zero.

49. Mag. field lines are continuous closed loops, but electric field lines are discontinuous.

50. Soft iron is used to make core of transformer, hysterisis loss is minimum for soft iron.

*Wd ? work done

*EF ? electric field

my passion...

Electrostatics Examples

You are advised to review the following examples. You should study them very closely and fully understand how they work. They all come from the preceding course, but the concepts will be used also in the present course.

The problems have been laid out as follows:

EG1 - forces and fields for single point charge
EG2 - forces and fields for two point charges
EG3 - 1-d motion in an electric field
EG4 - 2-d motion in an electric field

EG 1

Case

Problem involving forces, electric potential energy, electric field and electric potential for two interacting point charges or charged spheres.

Statement

A charge q₁ of - 2 x 10^{- 6} C is placed 15 cm from a charge q₂ of + 3 x 10^{- 6} C. What are:

The force on q₂.
The potential energy of the system of two charges.
The electric field due to q₁ at the point P where the charge q₂ is placed.
The electric potential due to q₁ at the point P where the charge q₂ is placed.

Preliminaries

Draw a figure and label it. Since directions are involved, it is probably best to lay it out on a cartesian co-ordinate system (i.e. put in some x and y axes). I have chose to put q₁ at the origin. Then work out intuitively the direction of the force on q₂

We immediately know that:

The force on q₂ will be towards q₁ or attractive (the two charges have different sign). On the figure that means in the negative x direction.
The potential energy the system of two charges will be negative (the two charges have different signs).
The electric field at P will be towards q₁ (the field will be in the direction of the force on a positive charge and a positive charge will be attracted to q₁) or in the negative direction.
The electric potential at P will be negative (the electric potential is related to the electric potential energy of a positive test charge and that will be negative if that charge is attracted).

Calculations and Answers

The electrostatic force between two charged particles is given by Coulombs Law:

For our problem the direction of (away from the charge q₁ towards the charge where we need the force) is in the positive x direction. Therefore, write instead of .

Put in the numbers:

Answer: The force on q₂ is 2.40 newtons in the negative x direction (that is, towards q₁). The direction is consistent with our preliminary analysis (q₂ is attracted to q₁).
Recall that the potential energy for the interaction of two point particles is given by:

Put in the numbers:

Answer: - .359 N*m. The negative sign is consistent with our preliminary analysis.
Electric field at P due to q₁. Remember from Coulomb's Law that the force on a test charge q_o at P is:

The electric field at P is defined to be:

Using the numbers given, we have:

Answer: The electric field at P (due to q₁) is 8.0 x 10⁵ newtons per coulomb or, equivalently, volts per meter in the negative x direction.
Electric potential at P due to q₁. You can obtain this via the electrostatic potential energy. The potential energy of a test charge q_o at P is given by:
The electric potential at P is thus:
From the given information:

Answer: The electric potential at P due to q₁ is - 1.2 x 10⁵ Volts or - 120 kV.

Comment
The problem was worked out as though the charges were points. For charged spheres, the procedures and answers are the same provided the radii of the spheres are less than the separation of their centers.

EG 2

Case
Forces and fields due to two (or more) charges.
Statement
Charge q₁ = - 2 x 10^{- 6} C is placed at x = 0 and y = 5 cm.
Charge q₂ = + 3 x 10^{- 6} C is placed at x = 0 and y = - 5 cm.
Charge q₃ = + 4 x 10^{- 6} C is placed at a point P which at x = 10 cm and y = 0 cm.

What is the net force on q₃?

What is the potential energy of q₃ due to its interaction with q₁ and q₂?

What is the electric field at P due to q₁ and q₂?

What is the electric potential at P due to q₁ and q₂?

Preliminaries
Draw a figure and label it. The problem statement has (very kindly) forced you to draw a co-ordinate system but if one had not been specified (e.g. if you had been given three charges at the points of some general triangle) then it would have been best to choose a co-ordinate system with one charge at the origin and as many as possible on the x and y axes. Once you've set up the co-ordinate system, draw the forces acting on q₃:

There are going to be two forces on q₃; they are vectors and will need to be added as vectors in order to give the net force on q₃. The direction of the resultant will obviously be upwards above the x-axis. I have drawn my vectors roughly proportional to the expected forces (the force due to q₁ will be less than that from q₂ since q₁ is smaller) and would guess that the resultant force is roughly 30 to 60 above the x-axis.
The potential energy of q₃ will be the (scalar) sum of the potential energies with q₁ and q₂. The potential energy from the interaction with q₁ will be negative (q₁ and q₃ have opposite sign) and that from the interaction with q₂ will be positive. The potential energy from q₁ will be the smallest. The scalar sum of the two potential energies will be small and positive.
Calculations and Answers

The electrostatic force on q₃.
We will need to work out each force separately and add them together as vectors. For clarity, redraw the above figure for each charge separately. Lets concentrate on q₁ first. Mark the angle between the x-axis and the straight line joining q₁ (causing the force) and q₃ (responding to the force).

Coulombs Law tells us that:
We are given q₁ and q₃, but do not know r₁₃ (the distance between q₁ and q₃) nor the unit vector . Let's sort those out first. The distance r₁₃ is worked out by geometry (Pythagoras' Theorem)

is a unit vector (which means its length is one) and makes an angle below the x-axis. Therefore, we may write it as:

Using this with Coulombs Law, we have:

Now, repeat the same steps for q₂. You should find:

(Notice that unit vector from q₂ to q₃ makes an angle above the x-axis; this explains the sign difference in the y-component of .)
To find the net force on q₃ we must add these two forces vectorially:

Answer: The net force on q₃ is N or 6.92 N at 68.2 above the x-axis.

Potential energy
Write the general form of the potential energy for two particles. Remember that potential energy is a scalar. Find the potential energy associated with each charge combination and add them (scalars add like "normal" numbers):

Answer: U = U₁₃ + U₂₃ = .322 J (or N*m).

Electric field at P due to q₁ and q₂
Work this out for each charge separately. The general expression for the electric field at a distance r away from a point charge is:

Draw a specific figure for q₁.

Work out the unit vector from q₁ to q₃ (it will be the same as in part (a)).
Write the complete equation for the electric field at P due to q₁:

Do the same thing for the electric field at P due to q₂:

Add the fields vectorially:

Answer: The electric field at P due to q₁ and q₂ is
(Note that the answer could also be obtained by taking the answer from part (a) and dividing by the charge q₃; convince yourself why that should be true!)

Electric potential
The general expression for the electric potential a distance r away from a point charge q is:

Write down the potential at P due to q₁:

Now find the potential at P due to q₂:

Add them:
V_net(P) = V₁(P) + V₂(P) = 8.10 x 10⁵ V.
Answer: The electric potential at P due to q₁ and q₂ is 8.10 x 10⁵ Volts or 810 kV.
(Note that the answer could also be obtained by taking the answer from part (b) above and dividing by the charge q₃; convince yourself why that should be true!)

Comments

If the charges were small spheres rather than points, the problem remains the same.

In part (b) we found the potential energy from the interaction of q₃ with q₁ and q₂. This is not the total potential energy of the whole system of three charges. To find the potential energy of the entire system:

Find the potential energy from q₁ interacting with q₂ (this is related to the work done in bringing q₂ in from infinity to its final position in the presence of q₁).

Find the potential energy from the interaction of q₃ with q₁ and q₂ together (which is related to the work done in bringing q₃ in from infinity in the presence of q₁ and q₂).

Add the results (this is related to the work done in bringing these charges into their final positions).

EG 3

Case
One-dimensional motion in an electric field.
Statement
You have a pair of large, flat parallel plates; one is positively charged and the other negatively charged. Both plates have the same charge density = 3 x 10^{- 8} C/m² (the only difference being the sign). The plates are separated by 10 cm.

What is the electric field between the plates?

What is the electric potential difference between the plates?

If a proton were released from the positive plate, what will its speed be just before it strikes the negative plate?

Preliminaries
Draw a figure and label it with the information that will be used and needed. It's best to use a side view:

Most of this problem involves calculations of quantities which have already been stated in the electrostatics notes.
What about part (c), i.e. finding the proton velocity? The proton in question is (of course) positively charged and will be in the electric field between the plates. Thus, it will accelerate and increase in speed as it is attracted towards the negatively charged plate. From our knowledge of the electric field, we can compute this force acting on the proton and therefore the proton acceleration. Once we know its acceleration, the velocity is obtained by the techniques learned in mechanics.
Solution and Answer

From the notes, the electric field between the two plates is simply twice the electric field due to just one of the plates:

From the notes, the potential difference between the plates is:
V₁ - V₂ = Ed = 3.39 x 10³ x 10^{- 1} = 339 V
Note: The positive plate is at the higher potential.

Particle Motion. First, lets use symbols; the force on a charge q in an electric field E is:

This is the only force acting on the charge (we're neglecting gravity; you can show that the gravitational force on a proton is orders of magnitude less than that of the electric force in this problem) so this must be equal to the mass of the particle times the particle's acceleration. Thus,

Notice that everything on the right-hand side of the equation is constant (remember, the electric field between the plates is uniform)! This tells us that the acceleration is constant! Recall what you learned about (one-dimensional) motion under constant acceleration from your mechanics course:

Getting rid of time between these two equations leads to:

In this case, the proton was released from rest so its initial velocity is zero; the acceleration was found in the previous part, so we have:

where d is the plate separation. Putting in the numbers, we calculate that the speed of the proton just before impact with the negative plate is 2.55 x 10⁵ m/s.
Answer: v_x = 2.55 x 10⁵ m/s or 255 km/s.

Comments
Is there another (perhaps easier) way to find the proton's velocity? Well, we know that energy is conserved; this means that the total energy of the proton will remain the same. Equating the proton's total energy when it begins its journey to just before hitting the negative plate, we have:
U₁ + K₁ = U₂ + K₂
U₁ - U₂ = K₂ - K₁
where K₁ is the kinetic energy and U₁ is the potential energy when at plate 1; K₂ and U₁ the values when at plate 2. K₁ is zero since the proton was initially at rest; U₁ - U₂ = q(V₁ - V₂) which we know from the relationship between the electric potential energy and the electric potential. So, we have:

Hence,

From which we find that v = 2.55 x 10⁵ m/s.
You could also work out how much time it takes for the proton to go from the positive plate to the negative. From the equations of motion and using the fact that the proton was initially at rest:

In this case, T = 7.85 x 10^{- 7} seconds (almost a microsecond; quite brief!!).

EG 4

Case
Two-dimensional motion in an electric field.
Statement
A pair of large, horizontal, parallel plates (one positively charged, the other negatively charged) produce a uniform electric field of 1.4 x 10⁶ V/m directed vertically downward. A liquid drop of mass 1.3 x 10^{- 10} kg with a positive charge of 1.5 x 10^{- 13} C enters the field initially moving horizontally (i.e. parallel to the plates) with a speed of 18 m/s. After the drop has moved 5 cm horizontally through the field, what will be the change in itsvertical position?
Preliminaries
This is basically how an ink-jet printer works! The drop could be a drop of ink and the electric field moves it in such a way that letters & pictures can be drawn on a sheet of paper. Should we worry about the force of gravity on the drop? The best strategy would be to leave that effect out at first, solve the problem simply as an electrical force problem, and then check whether gravity will make any difference.
Draw the plates and show the initial position and velocity of the drop:

Now, visualize where this drop will go. If there were no electric field, then it would continue straight across (remember, we're neglecting gravity) since there is a horizontal, but no vertical, velocity. What happens when there is an electric field? The field is directed downwards and the drop is positively charged; so, there will be adownwards force and thus a downwards acceleration. There is no force horizontally and therefore no acceleration horizontally. Thus, the horizontal velocity won't change yet the vertical velocity will. The vertical velocity, which is zero initially, will increase to some value, directed downwards, after the drop has moved a horizontal distance of 5 cm.
Draw that tragectory and label it; add sensible axes for convenience.

The game is to calculate the distance, y, that the drop has fallen after having moved a horizontal distance x.
The drop's acceleration can be worked out from the electrostatic force acting on it (see EG3); the solution is again like a mechanics problem. However, it may not be immediately clear what kinematic equations to use. A reasonable strategy would be to write down relationships between the position and time for both the horizontal and vertical directions and then see whether these relationships can be manipulated to give the desired answer.
Solution and Answer
From what we learned in mechanics, lets write out the relationship between position and time in the x direction:

where we've defined x_o to be zero (i.e. we've located the origin of the x-axis at the initial position of the drop). In this problem, there is no acceleration horizontally, so:

Write down the corresponding relationship between position and time in the y direction:

(and we've defined y_o to be zero; in other words, we've located the origin of our co-ordinate system at the initial position of the drop). In this problem, the initial velocity in the y direction is zero (that is, v_oy = 0)! Furthermore, we can relate the vertical acceleration to the electric field via:

where we've dropped the sub-script y on the electric field since its only directed vertically.
We know from our preliminary analysis that the acceleration must be directed down, so we must insert a negative sign on a_y (since we're taking upwards to be the positive y direction).
Therefore, the drop's vertical position as a function of time becomes:

So we have equations for the x and y co-ordinates as functions of time. We aren't given any information on how long it takes the drop to move 5 cm horizontally, so we'll need to eliminate time from these expressions:

Here is a relationship between the x and y co-ordinates; it happens to be a parabolic trajectory (which should be no surprise). Re-read the problem. We want to determine the value of y, i.e. the deflection, when x, the horizontal distance, is 5 cm. We know all of the quantities in the above equation!

When the drop has moved 5 cm horizontally through the electric field, the vertical deflection from the original (straight-line) trajectory is 6.23 mm downwards.

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1. What are we talking about?

The bottom line of this whole chapter is learning the difference between isomers. There are two types of isomers, constitutional and stereoisomers. Constitutional isomers are two compounds that have the same atoms present, but differ in their connectivity.
ie:

These compounds contain the same number of atoms, but the oxygen has been moved to form an ether instead of an alcohol. Therefore, these compounds are constitutional isomers.

Stereoisomers also have the same atoms present, however the connectivity is the same. This means the same number of hydrogens will be attached to each carbon and the same number of carbons will be attached to each carbon. Picture this:

Now, these structures both appear to be the same, but careful observation will reveal that the amine groups attached are in the cis conformation on the left and the trans conformation on the right. Therefore, the same atoms are present, but just in a different spatial arrangement.

Not to beat this idea into your head, but here is another example of a stereoisomer, but this time we will use a hydrocarbon chain.

Notice that the chain on the left is in the cis conformation at the double bond and the chain on the right is trans. This makes them stereoisomers.

2. I understand that chiral compounds are mirror images of each other that are not superposable, but how do I tell they are superposable?

The easiest way to tell if the mirror image is superimposable or not and superposable is to find the stereochemistry at the stereocenter. This entails you to find the stereocenter first and then label the groups attached to it in order of their priority. This means the atom with the highest atomic number will be labeled A and the next highest B. The next step is to rotate the molecule so the D group is facing away from you.
ie.

If the groups go from A to C clockwise, it is in the R configuration. If the groups are arranged counterclockwise, it is in the S configuration.

Practice a few

A B C

A has two stereocenters. The top stereocenter is an R configuration and the bottom stereocenter is an S configuration. For B the stereocenter is an S. C does not have to be considered because there are two of the same groups attached, and is not chiral.

If the two compounds you are looking at are mirror images of each other, but the configuration at the stereocenter differs, they are not superposable. Therefore they are chiral compounds. If they are superposable, then they are achiral.

3. How do I tell the difference between an Enantiomer and Diastereomer?

The easiest way to tell apart an enantiomer and a diastereomer is to look at whether or not the compounds are mirror images of each other. The best way to learn this is through practice. Here are a few examples, see if you can determine whether or not the compounds are enantiomers, the same, or diastereomers.

Hint: first determine if the compounds are mirror images of each other, and then find the individual stereochemistry around each chiral carbon. Remember the hand rule or the clockwise/counterclockwise arrangement discussed in the previous section.

D

If you are having problems determining the configuration at each stereocenter, I suggest building a model.

A is a pair of diastereomers, because the configuration is S, S in the first compound and R,S in the second compound.

B is a tricky one. They are both in the trans configuration and there is a plane of symmetry. Also, notice there is no carbon with four different groups. Therefore, they are not enantiomers and there is no stereochemistry.

C does not have a carbon with four different groups, so it does not have a stereocenter either.

D is a pair of enatiomers. Notice they are mirror images of each other.

4. There is an R and there is an S, but I don?t know what to do with them. Help!

If you have read the past few sections you know what the S and R designations are. They tell what type of stereochemistry is found at the stereocenter. Finding the stereochemistry at the stereocenters can help determine whether two compounds are enantiomers or diastereomers. Also, R and S versions of the same compound will have different optical activity values.

5. Quick Review of optical activity

Optical activity is the only physical property that differs from one enantiomer to the next. Optical activity is measured when plane polarized light is passed through a compound. When the light passes through the compound, it is bent either with positive rotation (dextrorotary) or with negative rotation (levorotary). There is no correlation between positive or negative rotation with the S or R configuration. S can be either dextrorotary or levorotary and the R enantiomer will be the opposite of the S. The value given to optical activity is specific rotation. The equation to figure out specific rotation can be found page 203 in your textbook.

6. Okay, I?m getting this stereocenter thing, but somebody had to go and screw everything up and stick two stereocenters together.

When dealing with two or more stereocenters on the same compound, there are a lot of possibilities. The first possibility is that the compounds are enantiomers of each other, the second that they are diastereomers, and finally that they can be meso compounds. Diastereomers occur when the compounds have the same chemical formula, but are not mirror images of each other.
ie.

Now look at these same atoms arranged differently to form an enatiomer. These compounds are mirror images of each other. However, they do have different stereochemistries, which makes them enantiomers.

You should also look at these next compounds and discover what makes them different from the above.

These compounds appear to be enatiomers, because they are mirror images of each other. They really are not. The middle two compounds are the meso compound, since they are the same. The outside two compounds are enatiomers of each other. Therefore, a meso compound is observed with stereoisomers where you would expect four different possible structures (two pairs of enantiomers), but there are only three stereoisomers.

7. Fischer Projections doesn?t mean a weekend out on the lake. How do I interpret them?

Fischer projections are a quick way to show three dimensions without the hassle of having to draw 3-D. They are very effective for those of us who lack artistic skills. When you look at the diagram the horizontal lines represent atoms that are coming out at you. The vertical lines mean they are going away from you. Fischer projections can be rotated 180 degrees and still be the same compound. However, if you flip it vertically or horizontally, it becomes the enantiomer.

This Fischer projection has been flipped horizontally. These two are enatiomers of each other. The first projection has an S, R configuration. The second projection has an R, S configuration.

Now lets look at a vertically flipped diagram.

These compounds are enatiomers of each other.

Finally, notice what happens when the diagrams are rotated 180 degrees in the plane of the paper.

The configuration at each stereocenter remains the same.

8. Cyclic Compounds

If you are anything like me, it is very hard for you to determine the stereochemistry in cyclic compounds the best way is just practice. Hopefully, this area will help. Do your best to determine the stereochemistry.

Analysis:

source;

facultystaff.vwc.edu/~jeaster/courseinfo/Tutorials/

congratulations!!!!

Electrostatics Examples

EG 1

EG 2

EG 3

EG 4

1. What are we talking about?

2. I understand that chiral compounds are mirror images of each other that are not superposable, but how do I tell they are superposable?

3. How do I tell the difference between an Enantiomer and Diastereomer?

4. There is an R and there is an S, but I don?t know what to do with them. Help!

5. Quick Review of optical activity

6. Okay, I?m getting this stereocenter thing, but somebody had to go and screw everything up and stick two stereocenters together.

7. Fischer Projections doesn?t mean a weekend out on the lake. How do I interpret them?

8. Cyclic Compounds

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