Chapters 22 and 23

Level I:

Note: When studying the following material, make sure to completely redraw the figures on your notebook and write the formulas as you proceed.  This will help you learn more.  Make sure to draw horizontal fraction bars when you use one and not a slash instead.

The Electric Charge and Electric Field:

In brief, electrons are  negative charges and protons are positive charges.  An electron is considered the smallest quantity of negative charge and a proton the smallest quantity of positive charge.

Two negative charges repel.  Also, two positive charges repel.   A positive charge and a negative charge attract each other (all experimentally verified).

The Point Charge:  An accumulation of electric charges at a point (a tiny volume in space) is called a point charge.   Defining a certain charge accumulation as a point charge is relative.  If the size of the charged particle is much smaller than the distances over which the effect of the charge is being studied, the charge accumulation is treated as a point charge.  For example, two charged sheets of A4 size paper that face each other and are 5" apart act as surface charges relative to each other.  The same two sheets of paper held 50 feet apart act as point charges relative to each other.

Note: When an atom loses an electron, the separated electron forms a negative charge, but the remaining that contains one less electron or consequently one more proton becomes a positive charge.  A positive charge is not necessarily a single proton.  In most cases, a positive charge is an atom that has lost one or more electrons.  In general, a particle is positive when many atoms of it have lost their electrons and negative when many atoms of it have excess electrons.  It is important to remind ourselves that a charged dust particle of mass 1pg (1 pico-gram = 10-12 gram = 1 trillionth of a gram) can easily have 1010 or 1011 atoms or molecules many of which could have lost electrons or gained excess electrons in a process.

The SI Unit of Electric Charge: Coulomb

If two like point charges are separated by 1m and the repulsion force between them is 9.0x109N, each charge is called 1 Coulomb, written as 1C.  It has been shown that  it takes 6.25x1018electrons to form 1C of negative charge or negative electricity.


Make sure you answer the questions or solve the problems before going any further For answers  click here.

Q1 It takes 6.25x1018 protons to form 1C of positive charge.  Find the charge of each proton. 

Q2.  How many electrons are there in -1mC, -1μC, -1nC, and -1pC of charge?{m = milli =10 -3, μ = micro =10 -6, n = nano =10 -9, and p = pico =10 -12}.

Q3.  Calculate the surface area of a sphere with a radius of (a) 1.00ft,  (b) 2.00ft, (c) 3.00ft, and (d) 4.00ft.   Spherical surface area is given by A = 4πr2.

Gold Leaf Electroscope:

An electroscope (See Fig. 1 below) is a device that detects the existence of electric charges on objects.  It is made of a glass compartment (such as a jar, for example) with a metal rod inserted into it through an insulator cap.  The end of the rod that is inside the compartment has two small metal foils (aluminum , gold, or another metal) hinged to it that are free to open up like the wings of a butterfly.  The outer end of the rod is connected to a tiny metal sphere or a pan.  When a charged object (no matter positive or negative) is brought into contact with the outer sphere or pan, some of the charges get transferred to the foils via the metal rod.  The foils become charged up with like charges and repel each other causing the foils to separate and open up.  That is how the foils indicate that some electric charges are transferred to them.  Even if a charged object is held near the sphere or the pan with no physical contact, the foils still open up, but if the object is taken away from the pan, the foils drop down again. Why?  Click on the following link and watch the process: . 

Figure 1

The Force Between two Point Charges:

The force between two point charges q1 and q2 separated a distance r from each other has a magnitude given by the

"Coulomb's Law:"    F = ( kq1q2)/r2   where  k = 8.99x109 Nm2/C2 is called the "Coulomb's Constant."

The direction of F is along the line that connects the centers of the two point charges as shown in Fig. 2.

Let's use red for positive charges and blue or purple for negative charges.

Figure 2

Note that the first two figures show that like charges repel.  The third figure shows that unlike charges attract.


Example 1:  Find the force between a 25.C charge and a 40.0μC charge when they are separated by a distance of 30.0 cm.  Both are point charges.

Solution:  F = (kq1q2 )/r2 ;  F = {9x109(25x10-6)(40x10-6) /0.32};   F = 100N directed away from each other. This applies to the first of the above 3 figures.


Q4.  Find the force between a -50.C charge and a -20.0μC charge when they are separated by a distance of 3.00 cm.  Let both be point charges.  First Solve. For answers   click here.

Q5.  Two balloons (basketball size) are connected by an East-West light thread and are positively charged.  What do you expect to observe if you cut the thread?  If the balloons were both negatively charged what would happen?

Q6.  If two metal spheres that have equal and opposite charges on them are brought into contact with each other, what will be the charge on each after contact?  Each sphere has an insulator mount. For answers  click here.

Q7. A metal sphere has +12μC of charge on it and another has -17μC.  If they are brought into contact, what will be the charge on each afterwards.  The spheres are identical and each has an insulator mount. 


Example 2:  In Fig. 3, find the force on charge q3.  Assume three significant figures on all quantities.

Solution: q1 attracts q3 in the direction shown by F13.    q2 repels  q3 in the direction shown by F23.

Figure 3

In SI, the magnitudes of F13 and F23 are:

F13  = kq1q3 /r132  = (9.0E9)(20E-6)(50E-6)/(52+52) = 9/50 = 0.180N at 135o.

F23  = kq2q3 /r232  = (9.0E9)(40E-6)(50E-6)/(52 +52) =18/50 = 0.360N at 45o.

Both angles are with respect to the positive x-axis.

The rest of the problem is what you learned in Physics I :  Vector Addition.

Rx = F13x + F23x   =   0.180cos(135o) + 0.360cos(45o)   =   0.127N.

Ry = F13y + F23y   =   0.180 sin(135o) + 0.360 sin(45o)   =   0.382N.


R = 0.403N,

θ = 71.6o.


Q8. Redraw the above example assuming all charges are positive.  Also, use the same y-distances of 5.0m, but change the x-distance to 8.0m.  Calculate (a) the angle that each of F13 and F23 makes with the positive x-axis.  (b) Find the magnitudes ofF13 and F23 .  (c) Determine the magnitude and direction of the resultant of F13 and F23 .  Note: If you feel you are not ready for this problem now, do it after you go through "Test Yourself 1", completely.   For answers   click here.

Test Yourself 1:

1) Like charges (a) repel   (b) attract   (c) neither a nor b.   click here.

2) Unlike charges (a) repel   (b) attract   (c) neither a nor b.   click here.

3) A charge is considered a point charge if (a) its dimensions with respect to the distances over which its effect is to be studied is quite smaller   (b) it has a zero diameter   (c) both a & b.

4) The force of a point charge on other equal charges around it that are also at the same distance has (a) the same magnitude and direction   (b) the same magnitude only   (c) different magnitude and different directions.   click here.

5) Charge +q1 is at (0,0) and +q2  at (5, 0).  The force of +q1 on +q2 points (a) West   ( b) East   (c)  North.

6) Charge +q1 is at (0,0) and +q2  at (5, 0).  The force of +q2 on +q1 points (a) West   ( b) East   (c)  North.

7) Charge +q1 is at (0,0) and +q2  at (0,- 4).   The force of +q1 on +q2  points (a) South   ( b) East   (c) North.

8) Charge +q1 is at (0,0) and +q2  at (0,- 4).   The force of +q2 on +q1  points (a) South   ( b) East   (c) North.

9) Charge -q1 is at (0,0) and +q2  at (- 4, 0).  The force of -q1 on +q2  points (a) South   ( b) East   (c) West.   click here.

10 ) Charge -q1 is at (0,0) and +q2  at (- 4, 0).  The force of +q2 on -q1  points (a) South   ( b) East   (c) West.

11) Charge +q1 is at (-3,0) and -q2  at (0, 3).  The force of  +q1 on -q2  points (a) Southwest   ( b) Northeast   (c) North.

12) Charge +q1 is at (-3,0) and -q2  at (0, 3).  The force of  -q2 on +q1  points (a) Southwest   ( b) Northeast   (c) South.

13) The angle of the force in Question 11 is (a) 45o   (b) 135o  (c) 225o.   click here.

14) The angle of the force in Question 12 is (a) 45o  (b) 135o  (c) -45o. 

15) Charge +q1 is at (-7,0) and +q2  at (0, 3).  The angle for the force of q1 on q2  is (a) 23.2o     (b) 203.2o    (c) -46.4o.

16) The distance between q1 and q2 in Question 15 is  (a) 6.32 units  (b) 7.62 units     (c) 5.62 units.   click here.

17) The distance between (0,5) & (5,0) is (a) 52 + 52 = 50 units  (b) (52+52)1/2 = 7.07 units    (c) 5 + 5 = 10 units.

18) The force of 25.0μC at (0,7.00m) on -12.0μC at (11.0m,0) is (a) 15.9mN, 32.5o    (b) 15.9mN, -32.5o    (c) 15.9mN, 147.5o.

19) The force of -45μC at (0,-4.0m) on 32μC at (9.0m,0) is (a) 0.13N, 204o     (b) 0.13mN, 24o     (c) 0.13N, 14o.   click here.

20) The force of -50.0μC at (-10.0m, 0) and 80.0μC at (10.0m, 0) on 20.0μC at (0, 0) is (a) 0.234N, 180o   (b) 0.234N,-180o (c) both a & b.  (d) neither a nor b.  click here.

21) The force of +50.0μC at (-10.0m, 0) and +50.0μC at (10.0m, 0) on 20.0μC at (0, 0) is (a) 0.180N, 180o   (b) 0.180N,-180o (c) zero.

22) The force of +50.0μC at (-10.0m, 0) and +50.0μC at (0, 10.0m) on 20.0μC at (0, 0) is (a) 0.127N, 45o   (b) 0.127N,-45o (c) 0.180N, 180o.   click here.

23)  The force of  +40.0μC at (0, -3.00m) and +20.0μC at (0, 3.00m) on 50.0μC at (4.00m, 0) is (a) 0.217N, 42o   (b) 0.217N,-42o (c) 0.891N, 14.0o.   click here.  If you learned how to do this problem, go back to Q8.


Electric Conductivity of Materials:

Classification of Electrons:

There are 3 types of electrons: bound electrons, valence electrons, and free electrons.

Bound electrons are the inner shells electrons in atoms that are under strong Coulomb forces from nucleus and difficult to detach.

Valence electrons are the outer shells electrons in atoms and participate in chemical reactions.  They are easier to remove from atoms.

Free electrons do not belong to any particular atom.  They flow in between atoms under the repulsive forces from the electron clouds of different neighbor atoms and the smaller attraction forces from the nuclei of the closest atoms.  Theconductivity of a substance depends on the number of free electrons that substance has.

Classification of Materials:

From the point view of conductivity,  materials are classified as conductors, semiconductors, and insulators.  The electric conductivity of a substance depends on its number or abundance of free electrons.

Metals are conductors.  Metals contain large number of free electrons per unit volume.

Nonmetals are insulators.  Nonmetals contain very few free electrons per unit volume compared to metals.

Semiconductors are alloys of metals and nonmetals.  They have controlled conduction properties depending on their metal percentages.

Static Electricity:

If electricity (the accumulation of negative or positive charges) can not flow easily, it causes localized charges and forms static electricity.  This happens when a bunch of electrons, for example, is given to a nonmetal ( insulator).  Because of lack of free electrons in the insulator, the transferred electrons stay locally where they are put and do not distribute in the insulator quickly.  They form static electricity. 

If a conductor (mounted on an insulator), is given a number of excess electrons, the electrons quickly distribute themselves in that conductor.  The insulator mount stops the electrons from flowing to the ground and becomes a boarder for them.  In the conductor part, due to the repulsion effect those excess electrons have on each other, they locate themselves as far away from each other as possible.  For a spherical conductor, the farthest possible distance is the outer surface of the sphere.  For other shape objects, it depends on their geometry. 

Fig 4 shows a metal sphere as well as an oval-shaped metal object, both on insulator mountings.  12 electrons are removed from the metal sphere and given to the oval metallic object.  The sphere becomes positive and the oval negative.  The charge distribution on the sphere is uniform due to symmetry.   Note the higher concentration of electrons (on the oval-shaped object) at the farthest possible distance that means the far ends.

Figure 4

Charging of an Object:

An object may be given electric charges in two ways:  1) by direct contact, and 2) by induction.

1) Charging by contact:

When a charged object is brought into direct contact with an uncharged (electrically neutral) object, part of its charges flow onto the uncharged object and makes it partially charged.  The transfer proportion depends on the shapes of the two objects.  For example, if the two objects are two identical metallic spheres with insulator mountings, they will divide the charges equally.  For asymmetric and unequal objects, the reasoning is more complicated and involved.  The three sections ofFig. 5 below show the simple case of two identical metallic spheres on insulator mountings a) before contact, b) during contact, and c) after separation.

Figure 5

2) Charging by Induction:

Charging by induction means charging without contact.  The Earth may be considered as being electrically neutral.  Adding a certain number of positive or negative charges to the Earth does not affect its neutrality.  Earth is so huge that the charges on the objects do not count at all compared to the charges that the Earth contains.  That is why Earth is electrically neutral for our experiments.  We can easily transfer some charges to it or take from it and it will not be affected.  If an electrically charged sphere (on an insulator mounting) is connected with a conductor (a metal wire) to the ground, it gets discharged either by transferring some electrons to the Earth or pulling some from it.   Fig. 6 shows how a positively charged sphere and a negatively charged one become discharged by being connected to the Earth.

Figure 6


Charging An Object Positively by Induction:   If a plastic rod is rubbed against wool, it becomes negatively charged.   If the rod is then brought close to a neutral metal sphere that is on an insulator mounting, it repels the free electrons of the sphere to the far end of it.  This makes the near end of it positive.  If the far end is connected to the Earth by a wire for a brief moment, the repelled electrons flow to the ground while the positive charges are held captive by the negative rod.  As soon as the connection with the ground is cut off, the rod may also be taken away leaving the sphere with the positive charges.  The process is shown in Fig. 7.

Figure 7

Charging an Object Negatively by Induction:  To be explained by

 students with appropriate figures.


The Electric Field (E):

Anywhere there is an electric charge q1, there exists the property of attraction or repulsion on other charges placed around it.  This effect of attraction or repulsion is called the electric field E1 of charge q1.  The electric field of charge q1 at Point P, depends on the amount of q1 and 1/r2 where r is the distance from the point charge.  We may come up with a formula for electric field E as

E1 is the magnitude of the electric field of charge q1 at Point P.  The direction of E1 is along the line that connects the center of the charge to Point P.

Figure 8

Using the statements made in the above two figures, we may write the Coulomb's formula in the following form:

Based on the above definition, the SI unit for electric field intensity is therefore, N/C.  The way the electric field strength E of a point charge q weakens with r is like the way light intensity weakens as we move away from a light bulb.  Suppose you  make a glass sphere with a surface area of 1ft2 with a tiny light bulb placed at its center.  Also suppose that you make another glass sphere with a radius twice as the first one and place it around the first sphere.  It is easy to show that when you double the radius of a sphere, its area quadruples.   If the two glass spheres are concentric, you can see why the light intensity at the surface of the outer sphere is 1/4 of the light intensity at the surface of the inner sphere.  You have already figured it out that the same amount of light energy that passes through the inner sphere must reach the outer sphere and pass through it as well.  Since the same energy is given to an area 4 times greater; therefore, the intensity of light on the outer sphere becomes 4 times weaker.  This is an example of  weakening as 1/r2.  What would happen to the light intensity (brightness) if you add glass spheres with radii 3x, 4x, 5x, 6x, and so on?  Again, as you have correctly visualized, the light intensity (brightness) would become 1/9, 1/16, 1/25, 1/36 and so on.


Figure 9Field Strength of a +q charge at different distances from it  

Fig. 9 shows how the field strength (shown by the vector length) of a positive charge +q weakens as the distance from the charge doubles and triples.  If a proton is placed once at r1=1, once at  r2=2, and once at  r3=3 units of distance from+q, the magnitude of  E3 at  r3=3 units becomes 1/9 of E1 at  r1=1 unit.  Also, the magnitude of  E2 at  r2=2 units becomes 1/4 of E1 at  r1=1 unit.  The same is true with a negative charge -q.  The only difference is that the field directions will be toward -q as shown in Fig 10.


      Fig. 10: Field Strength of a -q charge at different distances from it


The Test Charge:  One unit of positive charge is called the "test charge."  The best test charge is a single proton.  We use the test charge to examine the field strength generated by a point charge or a charge distribution at any desired point in space.  When a test charge is placed at different points around a point charge, it will be repelled or attracted by that point charge.  The farther away the test charge is placed, the weaker the force on it that means the weaker the electric field.  Note that the direction of the electric field is the same as the direction of the force on the test charge.  It is always along the line that connects the charge to the test charge.   The electric field direction is outward for a positive charge and inward for a negative charge as shown in Fig. 11.

Figure 11

Electric Field Lines Orientation:

In order to see how the electric field of a charge distribution varies from point to point in space, we may place the test charge at different points around the charge distribution and examine it. We will examine the field of 1) a single positive charge2) a single negative charge, and 3) an electric dipole.

For 1) a single positive charge and 2) a single negative charge, the field-lines orientations are already discussed and shown in Fig. 11, above.  Since a (+) charge repels away the test charge (a proton), its field-lines are drawn radially outward.  Also, since a (-) charge attracts the test charge, its field-lines are drawn radially inward.

3) The Field Lines Orientation of an Electric Dipole:

 Two equal but opposite charges separated by a distance L form the so-called an "electric dipole."  The field-lines orientation of a dipole is shown in Fig. 12.


Figure 12

Note: The meaning of each field line is as follows: if a test charge is placed on a field line, it will move on that curved line being constantly repelled from the (+q) and constantly attracted toward the (-q) until it is absorbed by (-q ).

The space around these two charges contains infinite number of points. Through each point there goes a field line.  Each point experiences the effects of two fields, one from the (+q) charge and one from the (-q) charge.    Two field vectors can be drawn at any given point:  an E+ that shows repulsion from +q, and an E- that shows attraction toward -q.  Of course, we are assuming that a test charge is placed at that given point at which the total field strength is being investigated.  This means that vector addition must be employed in order to find the resultant field. The following example clarifies the need for vector addition:

Example 3:  In Fig. 13, find the resultant electric field at Point A due to point charges q1 and q2.


Figure 13                                           Figure 14

Solution: In Fig. 13, Point A is an empty point in space.  If we place a test charge (a proton) at A, it will be repelled by q1 upward, but be attracted to q2 along the hypotenuse.  Such field vectors are shown in Fig. 14.  The goal is to first find the magnitudes of vectors E1 and E2, and then find their resultant.  The magnitudes are:

E1 = kq1/r12 ;   E1 = {9x109(120x10-9)/0.482}N/C ;   E1 = 4688N/C.

E2 = kq2/r22 ;   E2 = { 9x109(70x10-9)/(0.222 + 0.352 ) }N/C ;   E2 = 3686N/C.

Note that r22 = (hypotenuse)2 = 0.222 + 0.352 as in the denominator of E2.

The angle of E1 is 90o relative to the (+ x) axis; therefore,  E1 = (4700N/C, 90o).

The angle of E2 is α = tan-1(22/35) = 32.2o below the x-axis ; therefore, E2 = (3700N/C, -32.2o). 

 Vector Addition:

Ex = E1x + E2x = 4688cos(90o) + 3686cos(-32.2o) =  3119N/C.

Ey = E1y + E2y = 4688sin(90o) + 3686sin(-32.2o) =  2724N/C.

E = (Ex2 + Ey2)1/2 = {(31192 + 27242)1/2}N/C =  4100N/C.

θ = tan-1(2724/3119) = 41o.   Finally:     E = (4100N/C, 41o).

Example 4:  Calculate the electric field magnitude (strength) around a 25μC point charge at the following different distances: (a) 10.0cm, (b) 20.0cm, (c) 30.0cm, and (d) 50.0cm.

Solution: Left to students.

Uniform Electric Field:

   An electric field is called uniform if its strength does not change with distance.  The electric field of a point charge is not uniform because it strongly weakens as the distance from the charge increases.  It weakens proportional to 1/r 2 and its field lines diverge or open up very quickly in space.  There are other physical quantities that vary as 1/r 2.   Gravitational effect of mass M also weakens as 1/r 2.  Such quantities are said to follow the 1/r 2 law.   Is it possible to create an electric field that does not change with distance?  The answer is "yes".  If two parallel metallic plates are a certain distance apart and are connected to a battery, one plate accumulates negative charges while the other plate accumulates equal amount of positive charges.  The electric field in between the plates and away from the plates edges is essentially uniform and the electric field lines become parallel.  We will later on learn that such a device is called the "parallel-plates capacitor."   Fig. 15 shows the difference between the non-uniform field of a point charge and the uniform field of a parallel-plates capacitor.  The reason why the field lines are parallel is as follows:

   Each field line must emerge from the positive plate (and normal to it) and enter the negative plate (also normal to it).  Since the plates are parallel, the field lines become parallel as well.  The fact is that there are generally a large number of positive charges on the inner surface of the positive plate.    If each positive charge is to send out only one field line, there will be a very large number of field lines growing out of the positive plate that also repel each other. The result is necessarily parallel lines normal to the positive plate.  The negative plate that directly absorbs these lines, helps keeping them parallel.  The density of these parallel lines does not change in the space in between the plates making vector E, the electric field, a constant.  When a vector is constant, it is constant both in magnitude and direction.

Diverging (non-uniform) field lines (E varies with distance). 

Parallel (uniform) field lines (E is constant and does not change with distance).

Figure 15


Force-Field Relation:

If charge q is placed at a point where the electric field is E, it will experience force F such that

F = qE.

This may be concluded from Coulomb's law.   Coulomb's formula may be written as: 

F = kq1q2/r2   or    F = q2 (kq1/r2)      or      F = q2E1.

E1 is the electric field of charge q1.    q2 is in the field of q1.   

In general,

F = qE.

Example 5: In Fig. 16, an oil drop of mass 2.0μg (micrograms) that carries a charge of 450nC is placed at 5.0cm from the negative plate of a uniform electric field.  Find (a) the force on it by the 16000N/C field, (b) the work it does as it moves to reach the negative plate, (c) its P.E. at the beginning of its motion, (d) its K.E. just before hitting the negative plate, and (e) its speed just before hitting the negative plate.


(a) F = qE = 450x10-9C(16000N/C) = 7.2x10-3N.

(b) W = Fd = 7.2x10-3N(0.050m) = 360x10-6J.   

(c) P.E. = W = 360 μJ (Initial P.E. equals the work that it can do and did in Part (b).   

(d) The K.E. just before hitting the (-) plate is the same as initial P.E. ; thus,  K.E. = 360μJ as well.

(e) Since K.E. = (1/2)Mv2, we may solve for v to get:  v = (2K.E./M)1/2 = 6.0x102 m/s.


Figure 16

Example 6:  In Fig. 17two identical small balls 1.02grams each, are hung by two strings (1.00m each) from the same point.  When each is given a charge -q, they repel and the angle between each string and the vertical becomes 16o.  Find the amount of -q and its corresponding number of electrons.

Solution:  The distance between the two repelled balls r can be found by finding x and doubling it.  x is already calculated under the left figure; therefore, r = 0.551m. 

Figure 17

When the balls are given the same charge -q and the balls repel each other with force F, as shown in the right figure, the tension T in the cord becomes more than w.  Vector-wise, T balances w+F.  w+F is the dotted vector that is opposite to T. The angle between the dotted vector and w is also 16o.   We may write:

 tan16o = F/w;   F = w tan(16o) = Mg tan(16o) ;     F = 0.00287N.

Using Coulomb's formula, F = kq1q2/r2  with equal charges F = kq2/r2.

Solving for q2, we get:   q2 = Fr2/k ;   q2 = {0.00287(0.551)2/9x109 }C.

q = -3.1x10-7C;   # of electrons  = 3.1x10-7(6.25x1018 = 1.9x1012  = 1.9 trillions.



We will define another constant that will be used later.  This constant is εo, the "permittivity of vacuum" in allowing the electric field effect to pass through.  The symbol is pronounced "epsilon."   εo is related to k, the Coulomb's constant by the following relation:

εo = 1/(4πk).

The unit of εo is of course the reciprocal of that of k.   If you calculate its value, you will get:  εo =  8.85x10-12 C2/(Nm2).   The permittivity of air at normal pressure is not that different from that of vacuum.  This is because air at standard pressure is very dilute.  The permittivity of regular paper is 3.3 times that of vacuum, as an example.


Test Yourself 2:

1) Electrons are classified as (a) low Negative, medium negative, and high negative   (b) bound, valence, and free  (c) both a & b.  click here.

2) Bound electrons are (a) the outer shell ones   (b) the ones at the nucleus   (c) the inner shell ones.   click here

3) Free electrons are those that (a) do not belong to any particular atom and move in between atoms under the influence of repulsive forces from the electronic clouds of other atoms or the weaker attraction from the nuclei of the surrounding atoms   (b) are constantly being freed from the atoms   (c) neither a nor b.  click here.

4) Valence electrons are those that (a) may be freed from the outermost shells of atoms  (b) participate in chemical reactions   (c) both a & b.  click here.

5) Electrically, materials are classified as (a) conductors and insulators   (b) conductors, semiconductors, and insulators   (c) semiconductors and insulators.   click here.

6) Conductors contain a large # of free electrons per unit volume.  (a) True  (b) False  click here.

7) Semiconductors (a) are alloys of metals and nonmetals   (b) contain a controlled # of free electrons   (c) both a & b.

8) Insulators (a) contain a very low number of free electrons per unit volume  (b) are nonmetals   (c) both a & b.  click here.

9) The type of material that can best hold electric charges locally on its surface is (a) metal   (b) semiconductor   (c) nonmetal.

10) As soon as electric charges are given to a metal, (a) they stay locally where they are placed at   (b) they distribute throughout the metal object or might flow to the ground   (c) they always stay on the metal.  click here.

11) In order for electric charges given to a metal piece to stay on it (a) the metal piece must be on insulator mounting   (b) the metal piece must be on  insulator mounting but wired to the ground   (c) the metal piece must be initially neutral.  click here.

12) When electric charges are given to a metal sphere that is located on an insulator, the charges (a) flow to the lowest point of the sphere where the mounting is   (b) slowly distribute, but unevenly   (c) quickly distribute evenly over the sphere's surface.

13) A positively charged object is the one that (a) has protons distributed over its surface   (b) has more protons than electrons   (c) has lost a number of electrons   (d) b & c.  click here.

14) A metal sphere has 6.0μC and another identical one has 14.0μC of charge on it.  The charge on each after being brought into contact is (a) 4.0μC     (b) 0   (c)  10.0μC.   click here.

15) A metal sphere has 26.0μC and another identical one has -18.0μC on it.  The charge on each after being brought into contact is (a) 4.0μC   (b) 8.0μC   (c)  14.0μC.   click here.

16) A metal sphere has -20.0μC and another identical one has -10.0μC of charge on it.  The charge on each after being brought into contact is (a) -30.0μC   (b) +30.0μC   (c)  -15.0μC.   click here.

Problem: Two identical metal spheres A and B are on their insulator mountings, both initially neutral.   A is on the left and B on the right.  They are first brought into contact. A negatively charged  rod is then held close to the left of A.  To answer the following questions,  drawing a figure is absolutely necessary.

17) The rod makes the left of sphere A  (a) negative   (b) positive   (c) neutral.  click here.

18) The rod makes the right of sphere B  (a) negative   (b) positive   (c) neutral.  click here.

19) The reason why the right of sphere B becomes negative is that (a) the negative rod repels the free electrons in the connected spheres to the farthermost distance possible   (b) electrons cannot flow through both spheres   (c)  positive charges flow through both spheres to the left of A   (d) both a & c.  click here.

20) If you connect the right of sphere B to the neutral Earth, (a) the electrons flow to the Earth   (b) the Earth does not accept those electrons   (c) that end becomes positive.  click here.

21) If while the negative rod is still held near the left of A, the connection to the Earth of B is removed, (a) the left sphere is positive and the right one neutral   (b) both spheres are positive   (c) the left one is negative while the right one is positive.

22) If the negative rod is moved far away, (a) both spheres become negative   (b) both spheres become positive   (c) one sphere becomes negative and one positive.  click here.

23) The positive charges of the two spheres in contact stay (a) at the point where they are in contact   (b) at the left of each sphere   (c) equally at the left of A and the right of B.  click here.

24) If the spheres are separated, the charges on each (a) distribute themselves evenly on each sphere   (b) stay where they are   (c) fall toward the mounting on each sphere.

25) The electric field of a point charge (a) is uniform   (b) is non-uniform   (c) varies with 1/r.  click here.

26) The electric field of a point charge varies (a) with 1/r2   (b) with 1/r3   (a) does not vary with distance and is constant.

27) The electric field of charge q1 at a distance r from it is mathematically (a) E = kq1/r2   (b) E = kq1/r     (c) E = constant.

28) The way light from a light bulb weakens as we move away from it follows the (a) 1/r2 law   (b) 1/r law   (c) 1/r3 law.

29) According to 1/r2 law, doubling our distance from a light bulb, the light energy we receive at each eye becomes (a) 1/4   (b) 1/2    (c) neither a nor b.  click here.

30) According to 1/r2 law, quadrupling our distance from a light bulb, the light energy we receive at each eye becomes (a) 1/4   (b) 1/16    (c) 1/32.

31)  The electric field between two parallel plates oppositely charged (a) follows 1/r law   (b) follows 1/r2 law   (c) is constant and does not depend on the distance from either plate.  click here.

32) The coulomb's formula F = kq1q2/r2 may be written as (a) F = (kq1/r2) q2    (b) F = E1 q2 where E1 is the electric field of q1 at r   (c) both a & b.

33) According to the previous question, (a) Force = Field x Charge   (b)  F = Eq    (c) both a & b.  click here.

34) Two charges of q1 = 35μC and q2 = 45μC are placed at different distances from the negative plate of a parallel-plate capacitor (that has a uniform electric field in between its plates).  Draw a figure for it.  The field strength is 2000. N/Coul.  The force on the charges are: (a) .070N and .070N    (b) .070N and .090N     (c) .090N and .090N.

35) If the distance of q1 from the negative plate in the previous question is 4.0cm, the work done on q1 as it is pushed toward the negative plate by the field is (a) .0056J    (b) .0063J     (c).0028J.  click here.

36) Since .0028J of work is done by the field, we may say that the potential energy of q1 at 4.0cm from the negative plate is (a) .0028J    (b) .0028N      (c) .0028watts.

37) The energy of q1 as it speeds up toward the negative plate becomes more of  (a) elastic type   (b) potential type (c) K.E. type.  click here.

38) Due to energy conservation law, the K.E. of q1 just before striking the negative plate is (a) .0028N  (b) .0028J  (c) .0028w.

39) If q1 is on a mass of 12μgrams, its mass in kg is (a)12E-3 kg     (b)12E-6 kg     (c)12E-9 kg.  click here.

40) Knowing the K.E. of q1 just before hitting the negative plate, its speed is (a) 320m/s (b) 863m/s (c) 680m/s.


Level II:

Continuous Charge Distributions:

The formula E = kq/r2 in its vector form applies to point charges only.  To find the electric field of a certain charge distribution at Point P in space, the charge distribution must be divided into infinitesimal charges dq (so that each infinitesimal can be treated as point charge) and then the differential field (dE) of each (dq) be calculated at Point P.  The integral of such dE's will then give the field generated by that particular charge distribution at P.

Example 4:  The Electric Field Along a Uniformly Charged Slender Rod:

Find the electric field caused by a uniformly charged slender rod at Point P a distance a from one end of it.  The rod's length is L with a total charge of Q.

Solution:  Because of uniform charge distribution on this slender rod, if charge Q is divided by the rod's length L, we get the linear charge density λ = Q/L in units of Coul./m.  The infinitesimal charge that dx carries is therefore dq = λdx.  We treat each dq as a point charge and write the Coulomb's formula for it.


 The Electric Field Around an Infinite Line of Charge:

Example 5: Calculate the electric field intensity at a distance R from an infinite line of charge if its linear charge density is λ Coul./m.

Solution:  An infinitesimal length segment dl meters of the line carries a charge  dq = λdl  Coulombs.

The field that charge dq generates at P within dl    is dE = kdq/r2 that is

dE = kλdl /r2.  (dE is a vector).   (1)

For every dl  at a point like A, there is a dl  at a point like B where A and B are symmetric with respect to the line labeled R. Such symmetric dl's generate dEs that have equal and opposite y-components at P that cancel each other.  We end up adding only the x-components of dEs to come up with the total field at P.  Each x-component is

dEx = (dE) cosθ.      (dEx  is a scalar).

The magnitude of differential field at P becomes [from (1)]:

dEx = dl cosθ/r2.   (2)  Variables dl  and r2 must now be expressed in terms of θ.

The way dEx is completely expressed as a function of θ is shown below.


Example 6:  Find the electric field intensity E at point P that has a distance y from the center of a non-conducting disc of radius a if the surface charge density of the disc is s  Coul./m2.

Solution:  For every point on the ring, there is an opposite point on the other side of it that forms an equal and opposite x-component of dE at P.  Such x-components cancel each other and make ΣdEx = 0 at P.  Each y-comp. is

y = dEcosθ = (y/r)d  (4)

See if Equations (1), (2), and (3) on the right make any sense to you.  Note that dq is the charge on each ring (differential area).

If you substitute (2) in (1) and the result in (4), you will get (5).

Electric Field of a Dipole:

As was mentioned, an electric dipole is made of two equal and opposite charges that are distanced away.  Any molecule that has its center of positive charges away from its center of negative charges forms a dipole.  H2O and NaCl are examples.  Such molecules experience a torque on them when placed inside an external electric field causing them to rotate and align themselves along the direction of the external field.  To find the electric field of a dipole itself, let's place +Q and -Q at +a and -a on they-axis, and find the field they generate on the perpendicular bisector of the line that connects them, as shown.

The net field at A, a distance r from the origin is the resultant of E+ and E- as shown.  Since these two vectors have equal magnitudes, the resultant E becomes straight downward.  The x-components of E+ and E- cancel.   Their equal y-components are:

Torque on a Dipole: 

 When an electric dipole is inside a uniform electric field as shown, the external field exerts two equal but opposite forces on the dipole causing it to rotate and align itself along the field lines of the external field.  If the angle between the dipole moment P and the external field lines is θ, the magnitude of torque τ is

τ = 2F(d/2)sinθ = qE(d)sinθ = PEsinθ;  or,   τ = P x E (cross product)

where P = qd is the dipole moment.




1) Referring to Example 4, show that for a limited rod length, L, but large values of a, the derived formula reduces to E=kQ/a2.

2) Rework the integral of Example 5 to obtain an expression for E for a line of charge that is not infinitely long.  Let angle vary from -θ to +θ instead of from -π/2 to +π/2.

3) In the figure shown, (a) obtain an expression for the electric field at C if the rod segment is circular and uniformly charged with a linear charge density of l C/m, and (b) show that for a semicircular rod the field isE = 2kλ/R.

4)  In the figure shown, determine the minimum horizontal speed to the right at which an electron must be ejected near the negative plate such that it will not be absorbed by the positive plate before leaving the uniform electric field of 3600N/c that exists in between the plates.  Me=9.108x10-31kg    and    e=1.602x10-19C.

5)  In Example 6, Equation (7), what result do you obtain if you let the disk's radius, a, approach infinity?  In doing this, you are verifying that when the disk becomes an infinite plane of charge, the electric field becomes: E =σ /(2εo).  Note: σ is the lower case Greek symbol pronounced "sigma."

6) Find the electric field intensity in between two parallel and infinite sheets of charge with uniform surface charge densities ofσ and    by adding the electric field intensity of the individual fields.  Draw Part (c) on paper and draw the net electric field in Regions 1, 2, and 3.

7) In Example 6, Equation (7), what result do you obtain if you let the the distance of point P from the disk, y, becomes very large compared to a?

8) Find the electric field at a distance r from the center of an electric dipole if the point is along the line that connects the opposite charges.