Answer: See table above

i. Protons are positive; electrons are negative; more protons than electrons would mean an overall positive charge.

ii. Neutrons are neutral and will not have an influence on the overall charge. So this object could be +, -, or neutral depending on the relative number of protons and electrons.

iii. If a neutral object loses some electrons, then it will possess more protons (positive charge) than electrons (negative charge); the object will have an overall + charge.

iv. If a neutral object gains some electrons, then it will possess more electrons (negative charge) than protons (positive charge); the object will have an overall - charge.

v. A negatively charged object would never be attracted to a - balloon. However, a + object would be attracted to a - balloon (opposites attract) and a neutral object would be attracted to a - balloon (since neutral objects are attracted to any charged object).

vi. An object which attracts neutral paper bits must be charged. If the object also attracts a - balloon, then it must have a + charge (opposites attract).

vii. If an object has a repulsive interaction with any other object, then it is definitely charged. Since like charges repel, the charge which it has is the same type of charge as the charge on the object which it repels. This object has a - charge.

viii. A neutral object will be attracted to any charged object, whether + or -. In this case, the object attracts both + balloons and - balloons. Only a neutral object could do this.

ix. Balloon A is either + (since it attracts a - object) or neutral (since neutral objects will also attract any charged object). If our object attracts balloon A, then it is either + or - (since such objects would attract balloon A if it were neutral) or even neutral (since it would attract balloon A if balloon A were +).

x. Balloon C must be - since it repels another - balloon (like charges repel). If our object attracts balloon C, then it is either + (since opposites attracts) or neutral (since neutral objects are attracted to any charged object).

xi. Balloon E must be + if it repels balloon F (like charges repel). If our object repels balloon E, then it to must have a + charge.

xii. By definition, the electric field vector points in the direction which a + test charge would be accelerated. So by extension, the E vector points towards - charges and away from + charges. This object must have a - charge.

xiii. Using the same reasoning as above, this object must have a + charge.

Answer: B

This is an example of charging by friction. Two neutral objects, when rubbed together, will transfer electrons between them. The object with the greatest electron affinity will acquire electrons from the other object; the two become charged oppositely. In this case, the hair acquires electrons from the comb, leaving the comb with a shortage of electrons and a + charge; the hair now has an excess of electrons and a - charge.

Answer: C

In charging by induction, a charged object is brought near an object without touching it. The presence of the charge object induces electron movement and a polarization of the object. Then conducting pathway to ground is established and electron movement occurs between the object and the ground. During the process, the charged object is never touched to the object being charged.

Answer: AC

In charging by friction, the two objects being rubbed acquire opposite types of charge as electrons are transferred between objects.

In charging by contact, both objects acquire the same type of charge.

In charging by induction (as described in the answer to Question #17), the objects acquire the opposite type of charge.

Answer: ABDFGHJKLMN

a. TRUE The E vector depends upon two factors - the charge of the object creating the field and the separation distance from that charge.

b. TRUE The E vector depends upon two factors - the charge of the object creating the field and the separation distance from that charge.

c. FALSE An object can be used to measure the strength of an electric field; such an object is called a test charge. The test charge encounters an electric force. The electric field is the ratio of the force divided by the charge on the test charge. If the quantity of charge on the test charge is increased, the force is increased proportionately but the ratio remains the same.

d. TRUE An object can be used to measure the strength of an electric field; such an object is called a test charge. The test charge encounters an electric force. The electric field is the ratio of the force divided by the charge on the test charge.

e. FALSE The electric field strength is inversely proportional to the square of the distance of separation from the charge. If the distance is increased by a factor of 2, then the E is decreased by a factor of 4.

f. TRUE The electric field strength is directly proportional to the quantity of charge on the object which creates the field. Doubling the charge (Q) will double the electric field strength.

g. TRUE The electric field strength is directly proportional to the quantity of charge on the object which creates the field and inversely proportional to the square of the distance of separation from that charge. Doubling the charge will have the effect of increasing the E by a factor of 2; doubling the distance will have the effect of decreasing the E by a factor of 4. The combined effect of these two factors will result in decreasing the E by a factor of 2.

h. TRUE The electric field strength (E) created by object A does not depend upon the charge on the object used to test it. The electric field force (F) would be doubled by the doubling of the charge. But the electric field strength (E) is the force per charge - twice the force on twice the charge yields the same electric field strength.

i. FALSE The electric field strength created by object A does depend on how far from object A that the test charge is. Twice the separation distance means one-fourth the electric force (the inverse square law) and therefore one-fourth of the electric field.

j. TRUE This is a unique property of the electric field inside a closed conductor.

k. TRUE This is another unique property of the electric field. This property explains how lightning rods work. Being pointed, the electric field strength is large around the points (parts of highest curvature).

l. TRUE This is the customary convention used to define the direction of the electric field around any charged object.

m. TRUE Because electric field direction is defined as the direction which a positively-charged object would accelerate, the direction would be toward negatively charged objects (+ test charges are attracted) and away from positively-charged objects (+ test charges are repelled).

n. TRUE Since electric field is the force per charge exerted upon a test charge, the units would be force units per charge units - Newtons/Coulomb.

Answer: See table above.

I. The presence of the negatively-charged balloon repels electrons which are present in can G. Many of these electrons move into can H. This leaves can G charged positively (it lost electrons) and can H charged negatively (it gained electrons).

II. The presence of the negatively-charged balloon next to can J repels electrons which are present in can J. Many of these electrons move into can I. Similarly, the presence of the positively-charged balloon next to can I will attract electrons present in both cans. This combined effect leaves can J to become charged positively (it lost electrons) and can I to become charged negatively (it gained electrons).

III. The presence of negatively-charged balloons on each end of the collection of cans causes electrons to be repelled from the balloons. Many electrons leave the two cans on the ends and enter the can in the middle in order to distant themselves from the balloons. This leaves Can L charged negatively (it gained electrons) and Cans K and M charged positively (they lost electrons).

IV. The presence of the positively-charged balloon will attract electrons towards itself. There is a migration of electrons from the ground (hand) towards the balloon. This leaves can N charged negatively.

V. Electrons in can P will be attracted to the positively-charged balloon. This will polarize the can. But unless there is a source of electrons connected to can P, there is no way for the can to develop an overall charge. The can remains neutral.

Answer: F

Any explanation which involves the movement of protons can be quickly ruled out since protons are bound in the nucleus and incapable of moving about during ordinary electrostatic experiments. Electrons cannot move from the balloon to the pop can (nor vice versa) since there is no connecting path between these two objects. Air is a relatively good insulator and prevents the movement of charge between these two objects. What happens is that electrons in the hand sense the electric pull of the balloon and are drawn towards it, entering the pop can and imparting a negative charge to it.

Answer: D

The molecules of the wooden cabinet are neutral (which rules out diagrams A, B and E) yet polarized. The molecules polarize in such a way so as to make its forward side most appealing to the balloon. This means that the positive poles of the wood molecules are nearest the negative charge of the balloon. Only diagram D shows this orientation.

Answer: See diagrams above.

The direction of an electric field is the direction which a positively charged test charge would be accelerated. The electric field lines point in this same direction. As such, electric field lines are always directed away from + charges and towards - charges (or infinity). From this principle, it can be reasoned that A is positive, B is negative, C is positive and D is negative.

Answer: See diagram above.

In b., the presence of the positively-charged balloon serves to polarize the sphere. Electrons in the sphere are drawn towards the positively-charged balloon. This leaves a separation of charge in the sphere, with the excess negative charge on the left and the excess positive charge on the right.

In c., the sphere is touched to ground and electrons from the ground are drawn upwards and into the right side of the sphere. It is the presence of the + charge on the right side of the sphere which attracts these electrons to itself. At this point the sphere is charged negatively.

In d., this excess negative charge is shown distributed on the left side of the sphere, still drawn towards the positively-charged balloon.

In e,. the balloon is pulled away and there is a movement of electrons about the sphere so that the excess negative charge is uniformly distributed.

Answer: See diagrams above

Electric field lines begin at either infinity or a + charge and are drawn either to infinity or to - charges. Avoid crossing lines. At the surface of charges, the lines should be directed radially inwards (for - charges) or radially outwards (for + charges).

Answer: See diagrams above.

In a, c, and d, the movement of electrons is governed by the principles that opposites attract and like-charged objects repel. Electrons (being negative) are attracted to positively charged objects and repelled by negatively charged objects.

In b, the more electron-loving substance (highest electron affinity) gains the electrons from the object with which it is rubbed.

Answer: +2.4 x 10^-13 C

The overall quantity and type of charge on an object is dependent upon the relative number of protons and electrons in the object. To determine the quantity of charge, multiply the charge of one electron by the number of excess or deficient electrons. The type of the charge is dependent upon whether there are more protons or more electrons present in the object. If there are more protons, then the charge is +; otherwise, assign it a - value.

Answer: See table above.

The electric force (F_elect) is computed using Coulomb's law:

where Q₁ and Q₂ represent the charges on the two objects, d represents the separation distance between the object's centers and k = 9x10⁹ N/m²/C². This equation can be rearranged to solve for any quantity in the equation. The substitutions and algebra are shown below.

a. F_elect = (9x10⁹ N/m²/C²)•(1.6 x 10^-13 C)•(1.6 x 10^-13 C)/(0.5 m)² = 9.2 x 10^-16 N

b. d = SQRT[k•Q₁•Q₂/F_elect] = SQRT[(9x10⁹ N/m²/C²)•(1.6 x 10^-7 C)•(1.6 x 10^-7 C)/(0.017 N)] = 0.12 N

c. F_elect = (9x10⁹ N/m²/C²)•(4.65 x 10^-6 C)•(7.28 x 10^-6 C)/(0.658 m)² = 0.704 N

d. Q₂ = (d²•F_elect )/(k•Q₁) = (0.553 m)²•(0.0827 N)/[(9x10⁹ N/m²/C²)•(3.21x10^-6 C)] = 8.75^-7 C = 0.875 µC

e. Q = SQRT[(d² •F_elect )/(k)] = SQRT[(1.50 m)•(4.00x10^-4 N)/(9x10⁹ N/m²/C²)] = 3.16 x 10^-7 C

Answer: See table above.

This problem requires you to use the equations as a guide to thinking about how an alteration in one variable would effect the electric force and the electric field. The unique set of units on charge and force mean that you will not be able to plug-and-chug (unless you somehow develop a set of conversion factors between these charge units and force units and the conventional Newtons and Coulombs).

a. The electric field is the force per charge on the test charge. E = F/q = 40 mN/0.40 bP = 100 mN/bP

b. Since electric field does not depend upon the charge of the test charge, the E value at this same distance (compared to row a) will be the same value - 100 mN/bP. With twice the test charge (compared to row a), there will be twice the force - F = 80 N. Now check your thinking to see if E = F/q for your answers; if it does, then you've done well.

c. Here is some complex thinking: the charge creating the field is the same as in row a and row b. But the electric field strength is one-fourth the value; so the distance must have changed. Twice the distance would result in one-fourth the E value. So the distance is 2.0 m. Since electric F also depends inversely upon the square of distance, the electric force must be one-fourth the value of row a (where the q_B was the same value).

d. Since the electric field is the force per charge on a test charge, the equation F = E•q can be written. Now substitute values for E and q to solve for F. It is possible to approach this differently without a plug-and-chug approach. Compare row d to row a. The distance is identical but the charge of object A is doubled; doubling the charge would result in a doubling of the force. That is, the new force should be twice the original 40 N value of row a.

e. Compare row e to row d. The only difference between these two rows is the distance in row e is two times row d. Both electric force (F) and electric field (E) are inversely related to the square of the distance. So a doubling of d will result in a fourfold decrease in the value of both F and E. Divide the row d values by 4 to determine the row e values.

f. Once more, compare row f to row d. The only difference between these two rows is the distance in row f is one-half row d. Both electric force (F) and electric field (E) are inversely related to the square of the distance. So a halving of d will result in a fourfold increase in the value of both F and E. Multiply the row d values by 4 to determine the row f values.

g. Now compare row g to row d. In row g, there is triple the test charge and triple the distance. Tripling test charge will triple F and tripling distance will make F one-ninth the original value. So to determine the electric force, take the original (i.e., row d) value of 80 N and multiply by 3 and divide by 9; the result is approximately 26.7 N. To determine E, recognize that alterations in the test charge does not effect the electric field. The electric field in row g will be one-ninth the value of row d - approximately 22.2 N. As a check, divide the force by the test charge to see if indeed it is the same as the electric field value.

Answer: See sentences above.

a. If the distance is doubled, then the electric force is one-fourth the original value: 0.320 N/4 = 0.0800 N.

b. If the distance is decreased by a factor of 2 (halved), then the electric force is four times the original value: 0.320 N * 4 = 1.28 N.

c. Decreasing the distance by a factor of 3 will increase the force by a factor of 9; increasing the charge by a factor of 2 will increase the force by a factor of 2. So take the original force adjust accordingly: 0.320 N * 9 * 2 = 5.76 N

d. Increasing the distance by a factor of 3 will decrease the force by a factor of 9; increasing both the charges by a factor of 2 will increase the force by a factor of 2*2. So take the original force adjust accordingly: 0.320 N / 9 * 2 * 2 = 0.142 N

e. If the distance is increased by a factor of 2 (from 0.200 m to 0.400 m) then the electric field strength is decreased by a factor of 4. So take the original E value and adjust accordingly: 0.050 N/C / 4 = 0.0125 N/C.

f. If the distance is increased by a factor of 5 (from 0.200 m to 1.000 m) then the electric field strength is decreased by a factor of 25. So take the original E value and adjust accordingly: 0.0500 N/C / 25 = 0.00200 N/C.

g. Altering the amount of charge on the test charge will not effect the electric field strength at the same location. So the E value remains the same.

h. Altering the amount of charge on the test charge will not effect the electric field strength at the same location. But decreasing the distance by a factor of 2 will increase the electric field strength by a factor of 4. So take the original E value and adjust accordingly: 0.0500 N/C * 4 = 0.200 N/C.

Answer: 310 cm

This problem involves determining a location along the axis where two vectors are directed in opposite direction and have the same magnitude. At such a location the electric field created by object A equals the electric field created by object B.

A diagram is always a useful way of approaching the problem. Since A is positively charged, the electric field created by A is directed away from it. Since B is negatively-charged, the electric field created by B is directed towards. Thus, some directional reasoning would lead to the conclusion that the location along the axis must be outside of the two charges. And since object B has the smaller charge and creates the smaller E field, the location will be closer to object B than to object A. That is, what object B lacks in terms of charge it must make up for in terms of distance in order to have an equal field to object A's. The results of this reasoning are shown in the diagram at the right.

The location of 0 electric field is a distance of x cm from the -16 µC charge and a distance of 62+x cm from the +25µC charge. These distances and charge values can be inserted into the equation above and algebra can be used to solve for x. (Note that the + and - sign has been dropped from the equations; directional meaning has been accounted for in the reasoning above.)

Divide each side through by k.

Take the square root of each side of the equation and then cross-multiply.

Since x represents the distance from object B and since object B is at a y-coordinate of 62 cm, the answer to the problem is 310 cm.

Answer: 0.21 N, left

Object A experiences two forces - the force of object B attracting it and pulling it leftward (F_BA) and the force of object C repelling it and pushing it leftward (F_CA). The diagram below depicts the situation and the forces.

The net electric force upon object A is simply the vector sum of these two forces. The strategy involves computing each force (F_BA and F_CA ) individually and then adding them together as vectors. The calculations of the two forces are shown below. (Note that the + and - signs have been dropped from the values on the charge since the directional information has already been accounted for in the analysis above.)

F_BA = k•Q_A•Q_B/d² = (9.0x10⁹ N/m²/C²)•(5.6x10^-6 C)•(4.2 x10^-6 C)/(1.2 m)² = 0.147 N, left

F_CA = k•Q_A•Q_C/d² = (9.0x10⁹ N/m²/C²)•(5.6x10^-6 C)•(7.7 x10^-6 C)/(2.4 m)² = 0.067 N, left

Now the net electric force on object A can be determined by adding the two forces resulting from the interactions with A and with B. The net force is 0.21 N, left. (0.214 N rounded to two significant digits)

Answer: 5.4 x10³ N/C, 293 degrees counter-clockwise from east

Both charged object A and charged object B create an electric field, the strength of which is dependent upon the distance from the charged object. This problem involves computing the electric field strength as created by both charges individually and as experienced at the origin (E_A and E_B). Since electric field strength is a vector, it can be added using rules of vector addition. An electric field vector always has a direction associated with the direction which a + test charge would be forced. Since object A is a negatively charged vector, the vector E_A is directed towards it (i.e., rightwards). Since object B is a positively charged vector, the vector E_B is directed away from it (i.e., downwards). Once computed individually, E_A and E_B will be added.

The diagram at the right depicts the situation. The calculations below represent the mathematics.

E_A = kQ_A/R_A² = (9.0x10⁹ N/m²/C²)•(5.8x10^-6 C)/(5.0 m)² = 2.088 x 10³ N/C

E_B = kQ_B/R_B² = (9.0x10⁹ N/m²/C²)•(8.9x10^-6 C)/(4.0 m)² = 5.006 x 10³ N/C

Now the two E values will be added as vectors. The sketch at the right depicts the addition. Since they are at right angles to each other, the Pythagorean theorem will be used to determine the magnitude and trigonometry will be used to determine the direction.

E_net = SQRT[(2.088 x 10³ N/C)² + (5.006 x 10³ N/C)²]

E_net = SQRT[2.942 x 10⁶ N²/C²] = 5.4 x10³ N/C

Theta = invtan(E_A/E_B) = invtan[(2.088 x 10³ N/C)/(5.006 x 10³ N/C)] = 22.6 deg.

So the direction of the vector is ~23 degrees to the right of down (east of south) or simply 293 degrees counter-clockwise from east.

Answer: 0.0070 N

This is simply a matter of using the definition of electric field - the electric field is the force per coulomb of charge. Expressed mathematically,

and rearranged, the force is given by the equation

By substitution,

Answer: approx. 141 electrons

a. This involves a conversion using Avogadro's number and the molar mass:

b. Ernie wishes to use the force of electrical attraction between the alpha particle and the electron bundle to balance the downward pull of gravity upon the alpha particle. So

The charge on the alpha particle can be determined from its composition - it contains 2 protons (and 2 neutral neutrons), each having a charge of 1.6x10^-19 C. So the charge of an alpha particle is 3.2x10^-19 C. The only unknown remaining is the charge on the electron bundle. Rearrangement of the equation yields the following expression for the charge on the electron bundle:

Substitution of known values into this equation yields 2.26x10^-17 C as the charge on the electron bundle. With each electron contributed 1.6x10^-19 C per electron, there must be approximately 141 electrons in the bundle.

Answer: 0.397 C

The balloon in this problem is experiencing an electric force resulting from the interaction with the Van de Graaff generator. Diagram A illustrates the situation. The amount of charge on the Van de Graaff generator will be related to the distance of separation (R), the electrical force of repulsion, and the charge on the balloon (which is given). So the problem-solving strategy involves determining the electrical force of repulsion using a force analysis and then using Coulomb's law to determine the charge on the Van de Graaff generator.

Diagram B above shows a free-body diagram for the balloon. The three forces must balance since the balloon is at rest. Therefore, the x-component of the tension force (F_x in Diagram C) must equal the electrical force (F_elect). And similarly, the y-component of the tension force (F_y in Diagram C) must equal the gravity force (F_grav). A trigonometric function relating the angle of 32.0 degrees to these components of forces can be written.

And due to the equivalency between the components and the other two forces, it can also be stated that:

The equation can be rearranged to solve for F_elect; the expression for F_grav and the related values can be substituted.

Now the separation distance (R) between the balloon and the generator can be determined. Observe in Diagram A that the distance R is related to the length of the string and the angle with the vertical. Using the sine function, R can be determined as follows:

Now with F_elect, R, and Q₁ known, Q₂ (the charge on the Van de Graaff generator) can be determined.

By rearranging, the equation becomes

Answer: See answers below.

a. How many electrons are in your body? It's kind of a neat question; and remarkably, a student of Chemistry and Physics knows enough to come up with an estimate.

The strategy involves first using the mass of the body, the percent mass values, the atomic mass values, and Avogadro's number to first determine the number of atoms of each element in the body; then use the atomic number of each element to determine the number of electrons contributed by each element (assuming each atom is neutral); finally add all the electrons to determine the total number of electrons. Here it goes:

When added, there are 24.3x10²⁷ electrons (and a presumed equal number of protons).

b. The force of attraction of these protons and electrons at a separation distance of 100 m can easily be calculated if the net + and net - charge is known. Simply multiply the number of electrons by the charge of one electron (1.6x10^-19 C); do similarly for the protons. The charge ends up being 3.9x10⁹ C. Now Coulomb's law can be used to compute the electric force of attraction: