Electric Potential Energy: Point Charges

In the following diagram, the central charge equals 10 µC. Surface A (rA = 3 meters) the potential equals: VA = kQ/r = (9 x 109)(10 x 10-6)/3 = 3 x 104 volts Surface B (rB = 1 meter) the potential equals: VB = kQ/r = (9 x 109)(10 x 10-6)/1 = 9 x 104 volts If a 2 nC charge were to be brought in from infinity and placed on surface A shown above, the amount of work done on the 2 nC charge would equal W = qΔV = 2 x 10-9(ΔV) 2 x 10-9(VA - V∞) 2 x 10-9(3 x 104 - 0) 6 x 10-5 J We say that the 2 nC charge has gained an electric potential energy of EPEA = 6 x 10-5 J.  By definition, the absolute potential at a position infinitely far from a point charge is defined to be zero. Similarly, the amount of work done on the 2 nC charge to bring it in from infinity and place it on surface B would equal W = qΔV = 2 x 10-9(ΔV) 2 x 10-9(VB - V∞) 2 x 10-9(9 x 104 - 0) 1.8 x 10-4 J We say that the 2 nC charge has gained an electric potential energy of EPEB = 1.8 x 10-4 J.  The difference between the 2 nC's electric potential energy at B and its electric potential energy at A would represent the work required to move the 2 nC change from a position on surface A to a position on surface B. Wdone by external force = q(ΔVabs) = ΔEPE In essence, positive voltage changes mean that an external agent must do work to move a positive charge to a new position in the field, while negative voltage changes mean that the field would be doing all of the work to move a positive charge to its new position in the field. By definition, charges flow naturally from points of high potential to points of low potential. That is, when free to move, a positive charge would instinctively flow from surface B (high potential) to surface A (lower potential). Because of this, work done by electric fields (that is, when the charge moves along a field line in the direction of the field) results in a charge LOSING electric potential energy - that is, the electrostatic force causes a charge to move to positions of lower potential and less electrical potential energy consequently gaining KE. An analogy can be formed between equipotential surfaces and altitudes from the surface of the Earth. At any given altitude, an object with mass has a certain amount of gravitational potential energy, PEg = mgh where h is measured from some arbitrarily set zero level (usually the base of the hill). An external agent must do work against the gravitational field whenever the object's height, altitude, is increased  - this results in the object gaining PEg. The gravitational field does work on the mass whenever the object's height, altitude, is decreased resulting in the object losing PEg. By comparing the aerial view with the side view, you can tell that when the surfaces are closer together on the left, it signifies that the altitude is changing more rapidly, that is, that the slope of the hill is steeper. But regardless of which side of the hill a person climbs, he will do the same amount of work and gain the same amount of PEg = mgh. Remember that this is the definition of a conservative field. If instead, the aerial view where to be considered to be a series of equipotential surfaces, then the electric field is stronger on the left side than on the right side since the same changes in voltage occur in a smaller distance on the left than on the right. However, regardless of which direction a charge is moved from one surface to another, the same amount of work is done, since the charge gains or loses the same amount of electrical potential energy, Workdone by electric field = - q(ΔVabs ) = ΔKE The fact that these changes are path independent signifies that an electric field is also a conservative field -- that is, the only thing that counts is a comparison of the ending position to the initial position, not the path taken between the two points. Remember, whenever an electric field does work on a charged particle, the particle loses electric potential energy and gains kinetic energy. This is analogous to gravitational fields. When gravity does work on a mass, it loses potential energy and gains kinetic energy. If a system contains more than one charge, then the EPE of the system is the sum or the EPE of each pair of charge. EPEsys = k Σqiqj/rij In the following collection of charges, if each charge lies at the corner of a square of side s, then what is the EPEsys? Note that: r12 = r23 = r34 = r14 = s r13 = r24 = s  There are 6 "pairs" in this diagram: q1q2 q1q4 q2q4 q1q3 q2q3 q3q4 The total electric potential energy of this system would equal = k Σqiqj/rij = k (q1q2/r12 + q1q3/r13 + q1q4/r14 + q2q3/r23 + q2q4/r24 + q3q4/r34) = k (-q2/s + -q2/s + -q2/s + q2/s + q2/s + q2/s + q2/s) = k ( - q2/s +  -q2/s + -q2/s +  q2/s +  q2/s +  q2/s + q2/s) = k (q2/s)

Shells and Conductors

What is a conductor?   Conductors are materials, for example, metals, through which charged particles move readily.   What is meant by the term "under electrostatic conditions?" Under electrostatic conditions, field lines must terminate or begin on the surface of a conductor - that is, there is no electric field within a conductor. If any field line penetrated into the conductor, then electrons would respond to its presence and be accelerated within the conductor. If that happened, the conductor would no longer be under electrostatic conditions. Under electrostatic conditions, field lines must meet the surface of a conductor at right angles. If any field line did NOT come it at a 90º angle, then a component of the field line would be parallel to the conductor’s surface and electrons would respond to its presence and be accelerated within the conductor. If that happened, the conductor would no longer be under electrostatic conditions. When a conductor is under electrostatic conditions, all charges (electrons) must be at rest. Don't forget that one coulomb of charge represents 6.25 x 1018 electrons.   Under electrostatic conditions, the entire conductor must be at the same potential, or voltage. If not, then charges would flow from points of high potential to points of lower potential. If that happened, the conductor would no longer be under electrostatic conditions.   Faraday's Ice-Pail Experiment Faraday started with a neutral metal ice pail (metal bucket) and an uncharged electroscope. He then suspended a positively charged metal ball into the ice pail, being careful to not touch the sides of the pail. The leaves of the electroscope diverged. Moreover, their degree of divergence was independent of the metal ball's exact location. Only when the metal ball was completely withdrawn did the leaves collapse back to their original position. Faraday noticed that if the metal ball was allowed to contact the inside surface of the ice pail, the leaves of the electroscope remained diverged. Afterwards, when he completely removed the ball from the inside of the ice pail, the leaves remained diverged. However, the metal ball was no longer charged. Since the leaves of the electroscope that was attached to the OUTSIDE of the pail did not move when the ball touched the inside of the pail, he concluded that the inner surface had just enough charge to neutralize the ball.   Conclusions: Faraday’s Ice Pail Experiment pg 330-331, Principles of Physics, Frederick Beuche, McGraw-Hill Book Company, New York, New York. 1988. A charged metal object suspended inside a neutral metal container INDUCES an equal but opposite charge on the inside of the container. When the charged metal object is touched to the inside of the of the container, the induced charge exactly neutralizes the excess charge on the object. When a charged object is placed within a metal container, an equal charge of the same sign is FORCED to the outer surface of the container. All of the charge on any metal object resides on its outer surface if a conducting path is provided so that the charge can move there. Remember that charges will flow between two positions as long as there is a potential difference between those positions. When the voltage has been equalized, all charges will cease to flow. Faraday Cage   An important consequence of this experiment is that electric fields can be shielded - that is, the outside of a conductor acts as a FARADAY CAGE. A closed metal surface, no matter what it's shape, will block out any external electric field lines. And, as long as there are no electric charges residing inside the metal cavity, the electric field in the interior will be ZERO everywhere. This is why you are safe inside your car or on an airplane during a lightning storm.   Electrical shielding is easily accomplished by surrounding the surface that you wish to shield with a conducting surface. The free charges on the conducting surface will arrange themselves in such a way as to insure that the electric field within the conductor equals zero. This is the reason why electrical components come in metal boxes, to shield them from outside electrical activity.   This principle of electrical shielding is an important distinction between electric fields and gravitational fields. Electric fields can be shielded since there are two (2) types of electric charges. However, gravitational fields CANNOT be shielded - the effects of the gravitational attraction between two objects can be felt through any and all intervening matter.

Conducting Shells     Consider the charged conducting sphere shown above. Since there are eight field lines illustrated, let's assume that its charge is +8 µC. When viewed from infinity, this charged sphere would look like a point charge with an electric field that agrees with the graph for E vs r shown above.   However, consider that this charged sphere could instead be constructed of a NEUTRAL thin conducting shell with a hidden positive point charge located at its center, as shown below.     According to the results of Faraday's Ice Pail Experiment, the positive point charge INDUCES an equal but opposite charge on the inside of the shell AND an equal but similar charge on the outside of the shell. Note that the shell remains neutral - eight field lines terminate on the surface of its inner shell and eight field lines originate on the surface of its outer shell. Remember that there would be NO field lines between the "inner and outer" surfaces of the conducting shell. The only field lines would be between the inside point charge and the shell's inner surface and outside of the shell's outer surface. All field lines should be symmetric and meet any equipotential surfaces at right angles. When viewed from infinitely far away, this configuration would look exactly like the original 8 µC charged sphere! Given below is a diagram of the electric fields for this conducting shell.     For the remainder of this lesson we will work some examples using conducting shells. In each case, the conducting shell is aqua in color and the point charge placed in its center is yellow in color.

Electric Potential: Point Charges

When charged particles are moved from one position in an electric field to another position, a new unit of measurement is needed. A volt represents the amount of work per unit charge required to move a charge between two positions in an electric field. If it takes 1 joule of work to move 1 coulomb of charge between two positions in an electric field, then those positions have a potential difference of 1 volt. Voltage is a scalar property of an electric field, it has no direction, only magnitude. In general,   1 volt = 1 joule / 1 coulomb Rearranging these units (1 joule = 1 coulomb x 1 volt) shows us that the amount of work done on a charge by an external agent as it is moved around an electric field is expressed as   Wexternal = qΔV   For a point charge the absolute potential of any position in its electric field can be calculated using the equation   Vabs = kQ/r   When the charge creating the field is positive, the voltage is positive; when the central charge is negative, the voltage is negative. As r grows larger and larger, that is, as r approaches infinity, the absolute potential is defined to be zero. You can almost think of the "voltage" as being an indicator of the "elevation of the terrain" surrounding a  point charge. The steeper the terrain, the faster the voltage changes from one location to another. The work done by an external agent can be envisioned as "pushing or pulling" a second charge up or down these changes in elevation.   voltage "profile" - charge voltage "profile" + charge

Refer to the following information for the next three questions. The central charge, Q, has a charge of 10 µC.         What is the potential at surface A where rA = 3 meters?   What is the potential at surface B where rB = 1 meter?   Which surface has the higher potential?

Surfaces which connect points that are at the same absolute potential, or voltage, are called  equipotential surfaces. In the diagram of the point charge shown in the previous example, two equipotential surfaces were labeled, A and B. Notice that equipotential surfaces meet field lines at right angles. The closer together two equipotential surfaces are to each other, the more rapid the change in voltage. This indicates a stronger electric field which is shown in the second diagram below by the fact that the field lines are grouped closer together on the left side than on the right.     Note that the electric field strength, E, can be measured in either the units V/m, or equivalently, in the unit N/C.   N/C = V / d       = (J/C) / m       = [(Nm)/C] / m       = N/C   The following two graphs compare the voltage around a positively charged conducting sphere and the electric field for a positively charged conducting sphere. Note that the electric field strength (E ∝ 1/r2) drops off more rapidly than does the voltage (V ∝ 1/r). Also notice that within a conducting sphere, the voltage remains constant in contrast to the fact that no electric field exists.   For a conducting sphere, V = kQ/r For a conducting sphere, E = kQ/r2   Remember that the electric field strength, E, is a vector quantity. You are required to state both its magnitude and its direction to completely describe it at any given location. If you are ever asked to calculate the net electric field in 2-dimensions, you should first take the x- and y-components of each field, add the components to determine the net Ex and net Ey, and then calculate the resultant field and its direction. Voltage, on the other hand, is a scalar quantity and can be added directly without considering components or directions.

Gauss' Law

Qualitatively Gauss' Law is often stated as, "the net number of flux lines out of any closed surface containing a charge is proportional to the net charge inside the surface" or

But what is meant by the term electric flux, Φ? These are electric field lines penetrating a surface. If a field line enters the surface its value can be thought of as -1 while a field line existing the surface can be thought of as +1. The unit for flux is a Nm²/C.

Isolated positive charges are sometimes called sources since all of their field lines begin within the surface while isolated negative charges are sometimes called sinks since all of their field lines enter the surface.

positive isolated charge "+16" flux lines arrows point out		negative isolated charge "-16" flux lines arrows point in

If the surface encloses a mixture of charges, the number of flux lines is equal to the net, or sum, of the field lines entering and/or exiting the surface.

8 field lines leave and 6 field lines enter
(notice that 8 field lines are always inside the surface and are not counted)

net flux = +2 telling us that the positive charge is larger

Remember that the relative strength of an electric field can be represented pictorially as a proportional number of field lines. If one charge has twice the magnitude of another, it would have two times as many field lines.

Dot Product

When calculating the magnitude of the electric flux passing through a surface, the formula is

where

• E is the magnitude of the electric field,
• A is the cross-sectional area of the plane, and
• θ is the angle measured between the electric field lines and the normal to the area (which is called the area vector).

Remember that dot products produce scalar answers. So your results will no longer have a direction, only magnitude. That is, you will not be asked to find the components of the number of flux lines

In the top diagram the angle between the field lines equals 0º, so Φ = EA cos(0) = EA, its maximum value.

In the middle diagram the angle between the field lines equals 45º, so Φ = EA cos(45) = E(0.707A).

In the final diagram the angle between the field lines equals 90º, so Φ = EA cos(90) = 0.

To use Gauss' Law to calculate the electric field in a region, we choose a convenient Gaussian surface whose "edges or sides" lie either perpendicular or parallel to the field lines emanating from the charged object. We will only be responsible for highly symmetrical objects: point charges, charged wires/rods/cylinders, and sheets/disks of charge. We will begin our study with point charges.

Point Charges

Notice that by taking our Gaussian surface to also be a sphere, the field lines will always pass perpendicular to its surface [θ = 90º and cos(90º) = 1] so we can write Gauss' Law as

Charged Plane

Suppose we now look at a uniformly charged plane; that is, a surface

Electric Fields: Point Charges

An  electric field is the region surrounding a charged particle, Q, where another charged particle with experience either a force of attraction or repulsion. For point charges, the electric field lines are radial, getting ever farther apart as you get farther from the point charge itself. These fields are NOT uniform, but are examples of inverse square fields

E = kQ/r².

Dimensional analysis reveals that the units on E, the electric field strength, should be

E = kQ /r²
     (Nm²/C²)(C)(1/m²)
      N/C

Intuitively, the electric field strength measures the amount of force, in newtons, experienced by a coulomb of charge when it is placed at a particular position within an electric field.

Electric field lines point in the direction in which a positive test charge would respond to the electrostatic force; that is, away from positive charges and towards negative charges. In the following diagram, Q is positive, since the field lines are pointing away from Q. If Q had been negative, then the field lines would have pointed towards Q. Note that field lines are NEVER allowed to cross each other.

Another property of field lines is that they terminate on the surface of a charge - they do not penetrate into the charge. Consequently, there is no electric field within a charged conductor under electrostatic conditions. This fact is illustrated in the diagram of E vs r where it shows that the magnitude of the electric field equals 0 between 0 and r.

The size of two charges can be compared by noting the relative number of field lines surrounding each one. If a second charge with only 8 fields lines was compared to the diagram provided above, then it would indicate that the second charge was only ½ as large, since 8 is half of 16.

Fields between oppositely charged particles are attractive and are elliptical in shape; while fields between similarly charged particles are repulsive and hyperbolic in shape.

oppositely charged particles left is positive; right is negative	similarly charged particles both are positive

Physlet Animation: Electric Field Around Two Point Charges

A convenient way to remember the properties of an electric field are to use analogies to gravitational fields. A gravitational field is the region surrounding a massive object in which another object with mass will experience a force of gravitational attraction. One important distinction between electrical fields and gravitational fields is that electrical fields can be both attractive and repulsive; whereas gravitational fields are only attractive. Subsequently, gravitational fields cannot be shielded.

Gravitational Forces	Electrostatic Forces
G = 6.67 x 10-11 Nm2/kg2 is VERY small gravity is a weak force inverse square force attractive only	k = 9 x 109 Nm2/C2 is VERY large electrostatic forces are strong inverse square force attractive and repulsive
Gravitational Field	Electrostatic Field (+ charge)
gravitational field strength, g g's vector nature points towards the center of the planet each surface represents a unique value for g g is measured in N/kg (or m/sec2)	electric field strength, E E's vector nature points away from a positive charge or towards a negative charge each surface represents a unique value for E E is measured in N/C

In the chart above, you can see that both fields are inverse square relationships. That the electric field strength, E, has the same configuration as the gravitational field strength, g. That in each case, the force experienced by a second object equals the product of either that object's mass times the gravitational field strength or that object's charge times the electrical field strength. The direction of the gravitational field is defined as the direction a second object with mass would be attracted; whereas the direction of an electrical field is defined as the direction a positive test charge would respond.

Electrostatics Fundamentals

A positive charge means that the object has lost electrons and is no longer electrically neutral. Each electron lost gives the particle a charge of +1.6 x 10-19 coulombs. Positive, or vitreous, charges are classically created by rubbing a glass rod with silk. The rod becomes positive (loses electrons); the silk become negative (gains electrons). Since electric charge is conserved, the system (glass rod and silk) maintains a net charge of 0. A negative charge means that the object has gained electrons. Each electron gained gives the particle an additional charge of -1.6 x 10-19 coulombs. Negative, or resinous, charges are classically created by rubbing a rubber rod with fur. The rod becomes negatively charged; the fur positively charged. By definition, negatively charged objects have more mass than an identical neutral object since each extra electron has a mass of 9.11 x 10-31 kg. Three modes of electrifying an object: friction, conduction and induction. Electrification by friction occurs when two surfaces are rubbed together. Examples of this were discussed above when a positive charge was created by rubbing glass with silk and a negative charge was created by rubbing rubber with fur. The following list details a larger portion of the triboelectric sequence. When any two substances shown in this list are rubbed together, the top one will become positively charged while the lower one will become negatively charged. The further apart the two substances are in the list, the greater the electrification.  + - Asbestos Fur (rabbit) Glass Mica Wool Quartz Fur (cat) Lead Silk Human skin, Aluminum Cotton Wood Amber Copper, Brass Rubber Sulfur Celluloid India rubber   Charging by conduction means that the charging rod actually touches the electroscope’s knob. Since there is contact, electrons from the knob would flow onto a positive rod or off of a negative rod. Charging by conduction leaves the electroscope with a residual charge IDENTICAL to that of the charging rod. Charging by induction means that the charging rod is brought close to the electroscope’s knob but NEVER touches it. If the electroscope is not grounded, it will remain neutral but be temporarily polarized while the charging rod is in the immediate vicinity. That is, a positive rod will induce the electrons in the scope to migrate to the knob. This redistribution of charge will result in the leaves of the scope being positively charged If the electroscope is grounded during induction, electrons will flow from the knob to the ground if the charging rod is NEGATIVE and electrons will flow onto the knob if the charging rod is POSITIVE. The net effect once the grounding wire is removed is that the electroscope will be left with a residual charge that is OPPOSITE to that of the charging rod. Suppose the positive rod is brought near to an insulator, for example, a piece of paper or a section of a wall. Since electrons are not free to move within an insulator, another process takes place which still results in the paper or wall becoming polarized. The particles in the insulator realign themselves - presenting an oppositely charged layer towards the charged rod. This process is illustrated below.  positively charged rod top surface "-" polarized molecules  within the insulator bottom surface "+" Coulomb's Law: calculating electrostatic forces between point charges Coulomb’s Law of Electrostatics states that where: F is the force measured in Newtons k is Coulomb's constant which equals 9 x 109 Nm2/C2 Q is the magnitude of each charge measured in coulombs r is the distance between the centers of the two charges This formula may only be used for point charges. That is, for isolated points of electric charge. It is an example of an inverse square law: if you double the distance between two charges, the force between them is reduced to 1/4th its original size. If F is negative, that means that the charges carry opposite "signs" -- that is, one is positive and the other is negative. The negative answer means that the point charges are attracting each other -- it does NOT mean that F is acting in a negative direction. If F is positive, that means that the charges carry the same "sign" -- that is, either both are positive or both are negative. A positive answer means that the point charges are repelling each other -- it does NOT mean that F is acting in a positive direction.

Refer to the following information for the next four questions. Two identical conducting spheres are initially charged and separated at a distance of 1 meter, as shown below. What was the initial electrostatic force between them? After being brought together, touched, and then re-separated, what is the new charge on each sphere? What is the magnitude of the new force between? If one sphere had originally had a charge of +46 µC and the other had a charge of -46 µC, how would your answers have changed?

Summary: Take a moment to compare the properties of gravitational forces with those of electrostatic forces.  Gravitational Forces Electrostatic Forces G = 6.67 x 10-11 Nm2/kg2 is VERY small gravity is a weak force inverse square force attractive only k = 9 x 109 Nm2/C2 is VERY large electrostatic forces are strong inverse square force