The Electric Field

.. .. .. .. ..

The Electric Field Concepts and Principles Electric Charge, Electric Field and a Goofy Analogy We all know that electrons and protons have electric charge. But what is electric charge and what does it mean for a particle, like a neutron, to not have electric charge1? On one level, the answer is that electric charge is the ability to create and interact with an electric field. Of course, this begs the question, what is an electric field? To try to answer this question, let’s look at mass and the gravitational field. In Newton’s theory of gravitation, every object that has mass creates a gravitational field. The object with mass is termed the source of the gravitational field. The source’s gravitational field, which fills all of space, encodes information as to the location and mass of the source into space itself. It’s almost as if an infinite number of little business cards have been printed and distributed throughout space with detailed information concerning the source’s characteristics. The gravitational field can be thought of as a huge number of business cards embedded into the fabric of space. (No, I am not joking, it looks like this.)

1 kg source down there 1 kg source right here

1 kg source over there

1 kg source up there

1 kg source way over there

1

Actually, neutrons are constructed from smaller particles called quarks that do have electric charge. Neutrons are neutral because the sum of the quark charges inside of them is zero. The distinction between an object having a net charge of zero and an object having no charge whatsoever is important when considering polarization.

1

Typically when we draw the gravitational field we translate the message on the business cards into a mathematically equivalent message. Here’s a picture of the gravitational field near the surface of the earth.

9.8 N/kg

9.8 N/kg 9.8 N/kg

9.8 N/kg 9.8 N/kg

9.8 N/kg

9.8 N/kg

9.8 N/kg

9.8 N/kg

The relationship that allows you to “translate the business cards” is Newton’s expression for the gravitational field surrounding a mass,

 Gm g   2 rˆ r where 

G is the gravitational constant, equal to 6.67 x 10-11 N m2/kg2.



m is the source mass, the mass that creates the gravitational field.



r is the distance between the source mass and the location of the business card.



rˆ is the unit vector that points from the source mass to the business card. Notice that the direction of the gravitational field, g, is opposite to this direction.

Every mass in the universe has a gravitational field described by this formula surrounding it. Additionally, every mass in the universe has the ability to “read the business cards” of every other mass. This transfer of information from one mass to another is the gravitational force that attracts masses together. (Once one mass reads another masses’ business card, the first mass feels a strange urge to go visit the second mass. The more enticing the business card (the larger the value of g) the stronger the urge.) Let’s drop the business card analogy and return to electric charges and electric fields. Basically, every object that has electric charge surrounds itself with an electric field given by a formula incredibly similar to the one for the gravitational field. How incredibly similar? Below is the formula for the electric field at a particular point in space, from a single source charge:

 kq E  2 rˆ r where 

k is the electrostatic constant, equal to 9.00 x 109 N m2/C2.



q is the source charge, the electric charge that creates the electric field. Its value can be positive or negative, and is measured in coulombs (C).

2



r is the distance between the source charge and the point of interest.



rˆ is the unit vector that points from the source charge to the point of interest

Every charged object is surrounded by a field given by this relationship. Every other charged object in the universe can “read” this field and will respond to its information by feeling an electric force. Objects without electric charge neither create nor interact with electric fields (they can’t read the business cards). In this chapter you’ll learn how to calculate the electric field produced by charged objects. In the next chapter, you’ll learn how the electric field can be “sensed” by other electric charges resulting in the electric force.

Charge and Charge Density Macroscopic objects are normally neutral (or very close to neutral) because they contain equal numbers of protons and electrons. All charged objects are charged because of either an excess or lack of electrons. (It’s much easier to add or remove electrons from an object than trying to add or remove the protons tightly bound inside the nuclei of its atoms.) Thus, the electric charge of any object is always an integer multiple of the electric charge on an electron. Because of its fundamental importance, the magnitude of the charge on an electron is termed the elementary charge and denoted by the symbol e. In a purely logical world, the charge on any object would be reported as a multiple of e. However, since the charge on a macroscopic system can be many multiples of e, a more user-friendly unit, the coulomb (C), is typically used to quantify electric charge. In this system,

e  1.6 x10 19 C Thus, you can consider the charge on an electron as an incredibly small fraction of a coulomb, or a coulomb of charge as an incredibly large number of electrons. In many applications, in addition to knowing the total charge on an object you will need to know how the charge is distributed. The distribution of charge on an object can be defined in several different ways. For objects such as wires or other thin cylinders, a linear charge density, , will often be defined. This is the amount of charge per unit length of the object. If the charge is uniformly distributed, this is simply



Q L

where Q is the total charge on the object2 and L its total length. However, if the charge density varies over the length of the object, its value at any point must be defined as the ratio of the charge on a differential element at that location to the length of the element:



2

dq dL

I will always use uppercase Q to designate the total charge distributed on a macroscopic object and lowercase q to designate either an unknown charge or the charge on a point particle.

3

For objects such as flat plates or the surfaces of cylinders and spheres, a surface charge density, , can be defined. This is the amount of charge per unit area of the object. If the charge is uniformly distributed, this is



Q Area

or if the charge density varies over the surface:



dq dA

Lastly, for objects that have charge distributed throughout their volume, a volume charge density, , can be defined. This is the amount of charge per unit volume of the object. If the charge is uniformly distributed, this is



Q V

or if the charge density varies inside the object:



dq dV

To add to the confusion, you must realize that the same object can be described as having two different charge densities. For example, consider a plastic rod with charge distributed throughout its volume. Obviously, the charge per unit volume, , can be defined for this object. However, you can also define the object as having a linear charge density, , reporting the amount of charge present per meter of length. These two parameters will have different values but refer to exactly the same object.

4

Perfect Conductors and Perfect Insulators Determining how electric charges in real materials respond to electric fields is incredibly important but also incredibly complicated. In light of this, we will initially restrict ourselves to two types of hypothetical materials. In a perfect conductor, electric charges are free to move without any resistance to their motion. Metals provide a reasonable approximation to perfect conductors, although, of course, in a real metal a small amount of resistance to motion is present. When I refer to a material as a metal, we will approximate the metal as a perfect conductor. In a perfect insulator electric charges cannot move, regardless of the amount of force applied to them. Many materials act as insulators, but all real materials experience electrical breakdown if the forces acting on charges become so great that the charges begin to move. When I refer to an insulating material, like plastic, for example, we will approximate the material as a perfect insulator. Since electric fields create forces on electric charges, there cannot be static electric fields present inside perfect conductors. If a field was present inside a perfect conductor, the charges inside the conductor would feel an electric force and hence move in response to that force. They would continue to move until they redistributed themselves inside of the conductor in such a way as to cancel the electric field. The system could not reach equilibrium as long as an electric field was present. This re-arranging process would typically occur very quickly and we will always assume our analysis takes place after it is completed.

5

.. .. .. .. ..

The Electric Field Analysis Tools Point Charges Find the electric field at the indicated point. The charges are separated by a distance 4a. +2q

-q

The electric field at this point will be the vector sum of the electric field from the left charge (

  E L ) and the electric field from the right charge ( E R ).

Let’s examine the left charge first. Since the problem is expressed symbolically, the charge is simply “2q”, and the distance, r, between the charge and the indicated point of interest should be expressed in terms of “a”. Since each square in the diagram has width and height a, this distance can be expressed as:

r 2  (3a) 2  (a ) 2 r 2  10a 2

All that’s left to determine is the unit vector. The unit vector is simply a mathematical description of how to get from the source charge to the point of interest. In English, to get from the source charge to the point of interest you should move 3a in the x-direction and a in the y-direction3. This can be written as:

 r  3aiˆ  aˆj This is the vector that points from the source charge to the point of interest. However, this is not a unit vector since its magnitude isn’t 1. (A unit vector should convey a direction in space without altering the magnitude of the rest of the equation. It can accomplish this only if its magnitude is equal to 1.)

3

For the sake of consistency, we will use a common coordinate system with the +x-direction pointing to the right and the +y-direction pointing to the top of the page. For three dimensional systems, the +zdirection will point directly out of the page. The coordinate axes will be indicated in the vast majority of diagrams.

6

However, it’s simple to convert a regular vector into a unit vector, just divide the vector by its magnitude. This leads to:

rˆ  rˆ  rˆ  rˆ 

3aiˆ  aˆj (3a ) 2  a 2 3aiˆ  aˆj 10a 2 3aiˆ  aˆj 10a 3iˆ  ˆj 10

Putting this all together yields:

 kq E L  2 rˆ r  k (2q) 3iˆ  ˆj ) ( EL  10a 2 10  2 kq ˆ ˆ (3i  j ) EL  2 10 10 a  kq E L  0.063 2 (3iˆ  ˆj ) a  kq E L  (0.190iˆ  0.0632 ˆj ) 2 a

7

Repeating for the right charge gives:

 kq E R  2 rˆ r  k (q) (a )iˆ  (a ) ˆj ER  ( ) ( a ) 2  (a ) 2 (a ) 2  (a ) 2   kq ER  2 ( aiˆ  aˆj ) (a  a 2 ) 3 / 2   kq ER  (aiˆ  aˆj ) (2a 2 ) 3 / 2   kq E R  3 / 2 3 (aiˆ  aˆj ) 2 a  kq E R  0.354 2 (iˆ  ˆj ) a  kq E R  (0.354iˆ  0.354 ˆj ) 2 a

Adding these two contributions together yields

 kq E  (0.544iˆ  0.291 ˆj ) 2 a Thus, the electric field at the location indicated points to the right and slightly downward.

8

Continuous Charge Distribution The plastic rod of length L at right has uniform charge density . Find the electric field at all points on the x-axis to the right of the rod.

Electric charge is discrete, meaning it comes in integer multiples of electron and proton charge. Therefore, the electric field can always be calculated by summing the electric field from each of the electrons and protons that make up an object. However, macroscopic objects contain a lot of electrons and protons, so this summation has many, many terms:

 E



every electron and proton

kq rˆ r2

As described earlier, we will assume that the charge on macroscopic objects is continuous, and distributed throughout the object. Mathematically, this means we will replace a summation over a very large number of discrete charges with an integral over a hypothetically continuous distribution of charge. This leads to a relationship for the electric field at a particular point in space, from a continuous distribution of charge, of:

 k (dq ) E   2 rˆ r where dq is the charge on a infinitesimally small portion of the object, and the integral is over the entire physical object. Finding the electric field from a continuous distribution of charge involves several distinct steps. Until you become very comfortable setting up and evaluating electric field integrals, I would suggest you systematically walk through these steps. 1.

Carefully identify and label the location of the differential element on a diagram of the situation.

2.

Carefully identify and label the location of the point of interest on a diagram of the situation.

3.

Write an expression for dq, the charge on the differential element.

4.

Write an expression for r, the distance between the differential element and the point of interest.

5.

Write an expression for rˆ , the unit vector representing the direction from the element to the point of interest.

6.

Insert your expressions into the integral for the electric field.

7.

Carefully choose the limits of integration.

8.

Evaluate the integral.

I’ll demonstrate below each of these steps for the scenario under investigation.

9

1.

Carefully identify and label the location of the differential element on a diagram of the situation.

The differential element is a small (infinitesimal) piece of the object that we will treat like a point charge. The location of this differential element must be arbitrary, meaning it is not at a “special” location like the top, middle, or bottom of the rod. Its location must be represented by a variable, where this variable is the variable of integration and determines the limits of the integral. For this example, select the differential element to be located a distance “s” above the center of the rod. The length of this element is “ds”. (Later, you will select the limits of integration to go from –L/2 to +L/2 to allow this arbitrary element to “cover” the entire rod.)

2.

s

Carefully identify and label the location of the point of interest on a diagram of the situation.

You are interested in the electric field at all points along the x-axis to the right of the rod. Therefore, select an arbitrary location along the x-axis and label it with its location.

s x

3.

Write an expression for dq, the charge on the differential element.

The rod has a uniform charge density , meaning the amount of charge per unit length along the rod is constant. Since the differential element has a length ds, the total charge on this element (dq) is the product of the density and the length:

dq  ds

4.


By Pythagoras’ theorem, the distance between the differential element and the point of interest is:

r  x2  s2

10

5. Write an expression for rˆ , the unit vector representing the direction from the element to the point of interest.

The vector going from the element to the point of interest heads “down” a distance s and then to the right a distance x. This leads to a unit vector:

rˆ  rˆ 

6.

xiˆ  sˆj x 2  ( s) 2 xiˆ  sˆj x2  s2

Insert your expressions into the integral for the electric field.

 k (dq ) E   2 rˆ r  kds xiˆ  sˆj E 2 ( ) 2 x s x2  s2

7.

Carefully choose the limits of integration.

The limits of integration are determined by the range over which the differential element must be “moved” to cover the entire object. The location of the element must vary between the bottom of the rod (-L/2) and the top of the rod (+L/2) in order to include every part of the rod. The two ends of the rod form the two limits of integration.

 E

8.

kds xiˆ  sˆj ( ) 2 2  2 2 x  s x s L / 2 L/2

Evaluate the integral.

 E

kds xiˆ  sˆj (  2 2 x2  s2 ) L / 2 x  s L/2

L/2  E  k 

ds ( xiˆ  sˆj ) 2 3/ 2 L / 2 ( x  s ) 2

L/2   L/2  ds sds ˆj E  k  xiˆ   2 2 3/ 2 2 2 3/ 2   L / 2 ( x  s )  L / 2 ( x  s ) 

It’s typically easier to think of a vector integral as two, separate scalar integrals, one “in the xdirection” and one “in the y-direction”, as above.

11

Examining the x-integral (and using an integral table): L/2  E  kxiˆ 

ds 2 3/ 2 L / 2 ( x  s ) 2

L/2

   s E  kxiˆ   2 2 2  x x  s  L / 2  L E  kxiˆ x 2 x 2  ( L / 2) 2  kL E iˆ 1 x x 2  L2 4

I’ll leave it to you to evaluate the y-integral, but you should find that it equals zero. (Without doing any calculation whatsoever, you should know that the y-component of the electric field along the x-axis must be zero because of the symmetry of the situation. If it’s not clear why the y-component of the field must be zero, talk to your instructor!)

Thus, the electric field along the x-axis is given by:

 E

kL 1 x x 2  L2 4

iˆ

Does this function make sense? Two ways to check whether this field is reasonable are to determine how the function behaves for very small and very large values of x. As x gets very small, you are getting closer and closer to the charged rod. This should lead to an electric field that grows larger and larger (without bound). Notice that the limit of the function as x approaches zero does “go to infinity”, so the function does have the proper behavior for small x. As x gets very large, you are getting farther and farther to the charged rod. Not only should the field decrease, but as you get very far from the rod the rod should begin to look like a point charge. Notice that as x gets large, the term (L2/4) becomes negligible compared to x2. This leads to a field

 kL ˆ E i x x2  kL E  2 iˆ x

12

Since the total charge on the rod (Q) is simply the product of the charge density and the total length of the rod, this reduces to

 kQ E  2 iˆ x This is exactly the expression for the electric field from a point charge. Thus, as you move farther and farther from the rod, the rod does indeed begin to look like a point charge.

Gauss’ Law The long, hollow plastic cylinder at right has inner radius a, outer radius b, and uniform charge density . Find the electric field at all points in a plane perpendicular to the cylinder near its midpoint.

For certain situations, typically ones with a high degree of symmetry, Gauss’ Law allows you to calculate the electric field relatively easily. Gauss’ Law, mathematically, states:





 E  dA 

q enclosed

0

Let’s describe what this means in English. The left side of the equation involves the vector dot product between the electric field and an infinitesimally small area that is a piece of a larger closed surface (termed the gaussian surface). This dot product between electric field and area is termed electric flux, and is often visualized as the amount of field that “passes through” the little piece of area. The integral simply tells us to add up all of these infinitesimal electric fluxes to get the total flux through the entire closed surface. The gist of Gauss’ Law is that this total electric flux is exactly equal to the total amount of electric charge enclosed within the gaussian surface, divided by a constant, 0. (0 is the permittivity of free space, a constant equal to 8.85 x 10-12 C2/Nm2.) Somewhat counter intuitively, the key to applying Gauss’ Law is to choose a gaussian surface such that you never really have to do the integral on the left side of the equation! To try to help you understand what I’m talking about, let’s walk through the solution of the above problem. The following sequence of steps will help you understand the process of applying Gauss’ law: 1.

Choose the appropriate gaussian surface.

2.

Carefully draw the hypothetical gaussian surface at the location of interest.

3.

Carefully draw the electric field at all points on the gaussian surface.

4.

Write an expression for the surface area parallel to the electric field.

5.

Write an expression for qenclosed, the charge enclosed within the gaussian surface.

6.

Apply Gauss’ Law and determine the electric field at all points on this hypothetical surface.

13

Since we want to find the electric field at all points in space, there are three distinct regions we will have to investigate:  the region inside the “hole” in the cylinder (r < a),  the region within the actual material of the cylinder (a < r < b),  and the region outside of the cylinder (r > b). Let’s start outside of the cylinder.

Outside of the cylinder: r > b 1.


Gauss’ law is primarily useful when the objects under investigation have a high degree of symmetry. Gauss’ law basically exploits that symmetry to make the calculation of the electric field relatively painless. To exploit the symmetry of the situation, you should always choose a gaussian surface to mimic the symmetry of the object you are investigating. In this example, since the object is a cylinder, my gaussian surfaces will all be cylinders.

2.


Since we are trying to determine the electric field for all points outside of the cylinder, draw a cylindrical gaussian surface with radius r. The value of r is variable, and can take on any value greater than b, the radius of the real cylinder. Remember, the gaussian surface is hypothetical; it’s a mathematical “object” that only exists to help you solve the problem. Try not to confuse it with the real cylinder of radius b. The gaussian surface is represented from two separate viewpoints below. (The gray cylinder is the actual, charged cylinder while the dashed cylinder is the gaussian surface.) The gaussian surface has radius r, where r can be any value greater than b, and length L. It is located near the midpoint of the actual cylinder. The gaussian surface also includes the circular “end caps” of the cylinder since a gaussian surface must be a closed surface.

r

r

b L

14

3.


Although I have no idea what the magnitude of the electric field is at any point on my gaussian surface, the symmetry of the situation tells me that the direction of the electric field must be either radially away from or radially toward the central axis of the cylinder. If the charge density is positive, the field will be directed radially outward from the cylinder axis. Moreover, even though I don’t know the magnitude of the field, I do know that the magnitude is the same at every point on my surface.

E

4.

E


The left side of Gauss’ law requires us to evaluate an integral over the surface of our gaussian cylinder. The integral requires us to find the dot product between the electric field and the area, and integrate this dot product over the entire surface. I mentioned earlier that you should never have to actually do this integral (assuming you chose the “correct” gaussian surface). So why don’t we have to do this integral? The vector dot product can be re-written as:

  E  d A   ( E )( dA) cos  

where  is the angle between the electric field and the area of the gaussian surface. You may recall from calculus that the “direction” associated with area is perpendicular to its surface. Thus, the direction of each infinitesimal area is indicated on the diagrams below.

E

dA

15

E

dA

Notice that the electric field vector and the area vector are parallel at every point on the cylindrical portion of the gaussian surface and perpendicular at every point on the circular end caps. Thus, breaking the integral into two parts, one over the cylindrical portion of the gaussian surface and one over the end caps, yields:





 E  dA   ( E )(dA) cos   ( E )(dA) cos 0   ( E )(dA) cos 90   E  d A  

cylinder

endcaps

 ( E )(dA) cylinder

Now note that the magnitude of the electric field is the same at every point on the cylindrical portion of the gaussian surface since every point is equal distance from the charge distribution. Thus, the electric field is constant and can be brought outside of the integral, leaving a pretty easy integral to evaluate.





 E  dA   ( E )(dA) cylinder

   E  dA  E

 dA

cylinder

  E   dA  EAcylinder Notice that the entire left-hand side of Gauss’ law reduces to the product of the electric field magnitude and the magnitude of the surface area parallel to this field. Thus, because of our wise choice of gaussian surface, all we really need to calculate is the magnitude of the surface area parallel to the electric field. In this case, the parallel area is given by:

A parallel  Acylinder  2rL

5.


Since the gaussian surface is outside of the real cylinder, all of the charge on the cylinder within the length L is enclosed. Since we know the volume charge density on the cylinder, , the total charge on the cylinder within the length L is the product of the charge density and the volume of the cylinder.

q enclosed  V q enclosed   (Vsolidcylinder  Vhole ) q enclosed   (b 2 L  a 2 L)

16

6.


  qenclosed E   dA 

0 q E ( A parallel )  enclosed 0  (b 2 L  a 2 L) E (2rL)  0 E

 (b 2  a 2 ) 2 0 r

Thus, the electric field outside of the cylinder is inversely proportional to the distance from the cylinder. Now we have to repeat this analysis for the other two regions.

Within the cylinder: a < r < b 1.


We will again use a cylindrical surface. 2.


Since we are interested in the electric field within the actual cylinder, the radius of our gaussian surface is larger than a but less than b.

r a b

3.


As before, the magnitude of the electric field must be constant on the gaussian surface and directed radially outward due to the symmetry of the situation.

17

E

4.


The area parallel to the electric field is again the surface area of the cylindrical portion of the gaussian surface:

A parallel  Acylinder  2rL 5.


Since the gaussian surface is inside the real cylinder, not all of the charge on the real cylinder is enclosed by the gaussian surface. All of the charge between a and r is enclosed, but the charge between r and b is not enclosed by the gaussian surface. Thus, the amount enclosed is the product of the volume charge density and the volume of the portion of the cylinder enclosed by the gaussian surface.

q enclosed  V q enclosed   (Vwithingaussian  Vhole ) q enclosed   (r 2 L  a 2 L) 6.


E ( A parallel )  E (2rL)  E

qenclosed

0

 (r 2 L  a 2 L) 0

 (r 2  a 2 ) 2 0 r

Notice that this electric field increases with increasing r, since as r increases, more and more charge is available to produce the electric field. The last region we have to investigate is inside of the “hole” in the cylinder.

Inside the “hole”: r < a Since we must choose our gaussian surface to have a radius less than a, it is located inside the hollow center of the cylinder. Since there is no charge enclosed by this gaussian surface, the electric field in this region must be zero.

18

r

.. .. .. .. ..

The Electric Field Activities

19

Determine the direction of the net electric field at each of the indicated points. a.

+q

-q

+q

+q

b.

20

Determine the direction of the net electric field at each of the indicated points. a.

+q

+q

+q

b.

+q

+q

-q

-q

21

For each of the charge distributions below, indicate the approximate location(s), if any, where the electric field is zero. a.

+q

+q

+q

-q

+2q

-4q

+8q

-2q

b.

c.

d.

e.

+q

-q

22

+q

Two neutral metal spheres are insulated from the tabletop they rest on. A positively charged rod is initially far from the spheres. Before each action below, the system is returned to this initial condition.

A

++++

B

a. The spheres are near each other but do not touch, and the rod is brought near sphere A, but does not touch. The rod is held at rest near sphere A. Is the net charge on sphere A positive, negative, or zero? Zero Is the net charge on sphere B positive, negative, or zero? Zero Explain the reason for your response: Since the two spheres are electrically isolated from each other, the rod, and the tabletop, they must remain neutral. However, each sphere will be polarized by the presence of the charged rod. The left side of each sphere will become slightly negative and the right side slightly positive because some of the electrons in each sphere will move toward the positively-charged rod.

b. The spheres touch each other, and the rod is brought near sphere A, but does not touch. The rod is held at rest near sphere A. Is the net charge on sphere A positive, negative, or zero? Is the net charge on sphere B positive, negative, or zero? Explain the reason for your response:

c. The spheres are near each other but do not touch, and the rod touches sphere A. The rod is then removed. Is the net charge on sphere A positive, negative, or zero? Is the net charge on sphere B positive, negative, or zero? Explain the reason for your response:

d. The spheres touch each other, and the rod touches sphere A. The rod is then removed. Is the net charge on sphere A positive, negative, or zero? Is the net charge on sphere B positive, negative, or zero? Explain the reason for your response:

23

Two neutral metal spheres are insulated from the tabletop they rest on, but sphere B is grounded (connected to an infinite reservoir of mobile charges). A positively charged rod is initially far from the spheres. Before each action below, the system is returned to this initial condition.

A

++++

B

a. The spheres are near each other but do not touch, and the rod is brought near sphere A, but does not touch. The rod is held at rest near sphere A. Is the net charge on sphere A positive, negative, or zero? Is the net charge on sphere B positive, negative, or zero? Explain the reason for your response:

b. The spheres touch each other, and the rod is brought near sphere A, but does not touch. The rod is held at rest near sphere A. Is the net charge on sphere A positive, negative, or zero? Is the net charge on sphere B positive, negative, or zero? Explain the reason for your response:

c. The spheres are near each other but do not touch, and the rod touches sphere A. The rod is then removed. Is the net charge on sphere A positive, negative, or zero? Is the net charge on sphere B positive, negative, or zero? Explain the reason for your response:

d. The spheres touch each other, and the rod touches sphere A. The rod is then removed. Is the net charge on sphere A positive, negative, or zero? Is the net charge on sphere B positive, negative, or zero? Explain the reason for your response:

24

a. Four identical electric charges are equally spaced along a line. Rank the electric field created by these charges at each of the indicated points. Electric fields pointing to the right are positive.

A

B +q

C +q

D +q

E +q

Largest Positive 1. _____ 2. _____ 3. _____ 4. _____ 5. _____ Largest Negative _____ The ranking cannot be determined based on the information provided. Explain the reason for your ranking:

b. Four identical magnitude electric charges are equally spaced along a line. Rank the electric field created by these charges at each of the indicated points. Electric fields pointing to the right are positive.

A

B +q

C -q

D +q

E -q

Largest Positive 1. _____ 2. _____ 3. _____ 4. _____ 5. _____ Largest Negative _____ The ranking cannot be determined based on the information provided. Explain the reason for your ranking:

25

A solid metal sphere is centered between two metal spherical shells. The sphere has a charge +2Q and the shells are neutral.

A B C D E

The outer surface of the sphere. The inner surface of the inner spherical shell. The outer surface of the inner spherical shell. The inner surface of the outer spherical shell. The outer surface of the outer spherical shell.

a. Rank the surfaces on the basis of the charge on each surface. Largest Positive 1.ACE _____ 2.BD _____ 3. _____ 4. _____ 5. _____ Largest Negative _____ The ranking cannot be determined based on the information provided. Explain the reason for your ranking: The excess positive charge on the inner sphere will migrate to the outer surface of the sphere, because these positive charges will try to get as far away from each other as possible. This will draw negative charges on the inner spherical shell to the inner surface of the shell, until -2Q of charge is on this surface. Since this shell is neutral, this leaves behind +2Q on the shell which will migrate to its outer surface. This then draws -2Q onto the inner surface of the outer shell, leaving behind +2Q on the outer shell which again migrates to its outer surface.

b. Rank the surfaces on the basis of their charge density. Largest Positive 1. _____ 2. _____ 3. _____ 4. _____ 5. _____ Largest Negative _____ The ranking cannot be determined based on the information provided. Explain the reason for your ranking:

26

A solid metal sphere is centered between two metal spherical shells.

A B C D E

The outer surface of the sphere. The inner surface of the inner spherical shell. The outer surface of the inner spherical shell. The inner surface of the outer spherical shell. The outer surface of the outer spherical shell.

a. Each of the three objects has charge +2Q. Rank the surfaces on the basis of the charge on each surface. Largest Positive 1. _____ 2. _____ 3. _____ 4. _____ 5. _____ Largest Negative _____ The ranking cannot be determined based on the information provided. Explain the reason for your ranking:

b. The sphere has charge –3Q, the inner shell –2Q, and the outer shell +Q. Rank the surfaces on the basis of the charge on each surface. Largest Positive 1. _____ 2. _____ 3. _____ 4. _____ 5. _____ Largest Negative _____ The ranking cannot be determined based on the information provided. Explain the reason for your ranking:

27

The same total charge is uniformly distributed throughout the volume of four, identical rectangular insulators. Then one insulator has its top 1/3 (B) separated from its bottom 2/3 (C), one insulator has its left 2 /3 (D) separated from its right 1/3 (E), and the final insulator has its front 1/2 (F) separated from its back 1/2 (G).

B A

D

C

E

F

G

a. Rank the insulators on the basis of their total charge. Largest 1. _____ 2. _____ 3. _____ 4. _____ 5. _____ 6. _____ 7. _____

Smallest

_____ The ranking cannot be determined based on the information provided.

b. Rank the insulators on the basis of their volume charge density. Largest 1. _____ 2. _____ 3. _____ 4. _____ 5. _____ 6. _____ 7. _____

Smallest

_____ The ranking cannot be determined based on the information provided.

c. Rank the insulators on the basis of their linear charge density along the horizontal (x) axis. Largest 1. _____ 2. _____ 3. _____ 4. _____ 5. _____ 6. _____ 7. _____

Smallest

_____ The ranking cannot be determined based on the information provided. Explain the reason for your rankings:

28

a. For each of the scenarios described below, rank the objects on the basis of their net charge. All charge densities are uniform

A B C D E F

A 1.0 m long, 0.001 m radius copper wire with charge density 1.0 C/m. A 1.0 m long, 0.001 m radius copper wire with charge density 1.0 C/m2. A 0.5 m radius copper spherical shell with charge density 1.0 C/m2. A 2.0 m long, 0.1 m radius plastic cylinder with charge density 2.0 C/m3. A 0.5 m radius copper sphere with charge density 1.0 C/m2. A 2.0 m long, 0.1 m radius copper cylindrical shell with charge density 2.0 C/m2. Largest 1. _____ 2. _____ 3. _____ 4. _____ 5. _____ 6. _____ Smallest _____ The ranking cannot be determined based on the information provided.

Explain the reason for your ranking:

b. A charge Q is distributed uniformly throughout the volume of each of the plastic objects described below. Rank the objects on the basis of their charge density.

A B C D E F

A solid cube with edge length L. A solid cylinder of length 0.5L and radius 0.2L. A solid sphere of radius 0.2L. A solid half-sphere of radius 0.4L. A spherical shell of inner radius 0.2L and outer radius 0.4L. A cylindrical shell of length 0.5L, inner radius 0.1L and outer radius 0.2L

Largest 1. _____ 2. _____ 3. _____ 4. _____ 5. _____ 6. _____ Smallest _____ The ranking cannot be determined based on the information provided. Explain the reason for your ranking:

29

Find the electric field at each of the indicated points. The distance between the sodium ion (Na+) and chloride ion (Cl-) is 2.82 x 10-10 m. a.

Na+

b.

Qualitative Analysis On the graphic above, sketch the direction of the electric field at each of the indicated points. How should the electric field at pt. a compare to the electric field at pt. c? Explain.

Mathematical Analysis

30

Cl -

c.

Find the electric field at each of the indicated points. The distance between each sodium ion (Na+) and chloride ion (Cl-) is 2.82 x 10-10 m. Cl-

a.

Na +

b.



31

Cl-

c.

Na +

Find the electric field at each of the indicated points. The distance between each sodium ion (Na+) and chloride ion (Cl-) is 2.82 x 10-10 m.

Cla.

Na + b.

Na +



32

c. Cl-

Find the electric field at each of the indicated points. The charges are separated by a distance 4a.

a.

b. +q

d.


33

-q

c.

Find the electric field at each of the indicated points. The charges are separated by a distance 4a.

a.

b. +2q

d.


34

+q

c.

The two charges at right are separated by a distance 2a. Find the electric field at all points on the x-axis. +2q

-q

Mathematical Analysis for x > a: x

 2 1 k ( 2q ) ˆ k (  q ) ˆ ) E i i  kqiˆ(  2 2 2 ( x  a) ( x  a) ( x  a) ( x  a) 2

+2q

-q

Note that x is the position of the pt. of interest and although positive in this region will be negative in other regions. r is the distance btwn the pt. of interest and the charge and should always have a positive value.

x-a x+a

for -a < x < a:

 2 1 k ( 2q ) ˆ k (  q ) ( iˆ)  kqiˆ( ) E i  2 2 2 ( x  a) (a  x) ( x  a) ( x  a) 2 r for the right-hand charge is NOT (x – a) since this would return a negative number in this region. (Of course, since this term is squared it would not make a difference mathematically, but it is still conceptually incorrect.) The expression (a-x) returns a positive value for all values of x in this region.

x +2q

-q a-x

x+a

for x < -a:

 E

1 2 k ( 2q ) k (q ) (iˆ)  (iˆ)  kqiˆ( )  2 2 2 ( x  a) (a  x) ( x  a) ( x  a) 2

r for the right-hand charge is the sum of the magnitude of x and a. Since x is a negative number, the magnitude of x is –x, so the correct expression for r is –x + a, or a – x.

x +2q

-q

a + |x| = a - x |x| - a = -x - a

Sketch Ex below. E

a

-a

x

Note that the only point on the x-axis at which the field is zero is to the right of the negative charge. At all points btwn the charges the field is positive.

35

The two charges at right are separated by a distance 2a. Find the electric field at all points on the x-axis. +q

+2q

Mathematical Analysis for x > a:

for -a < x < a:

for x < -a:

Sketch Ex below. E

a

-a

36

x

The two charges at right are separated by a distance 2a. Find the electric field at all points on the y-axis. +q

-q


Questions For all points on the y-axis, what should Ey equal? Does your function agree with this observation?

Sketch Ex below. E

y

37

The two charges at right are separated by a distance 2a. Find the electric field at all points on the y-axis. +q

+q


Questions For all points on the y-axis, what should Ex equal? Does your function agree with this observation?

At y = 0, what should Ey equal? Does your function agree with this observation?

Sketch Ey below. E

y

38

Two charges are separated by a distance 2a. Determine the location(s), if any, where the electric field is zero. Mathematical Analysis a. +q

+2q x

The electric field is zero between the two charges, to the left of the midpoint. Labeling this position x results in:

 E

k (q) ˆ k ( 2q ) k ( q ) ˆ k ( 2q ) i (iˆ)  i (iˆ)  0 2 2 (a  x ) (a  x ) ( a  x) 2 ( a  x) 2 k (q) ˆ k ( 2q ) ˆ i i 2 (a  x) (a  x) 2 1 2  2 (a  x) (a  x) 2

(a  x)  2 (a  x) (1  2 )a  (1  2 ) x x

(1  2 ) (1  2 )

b. -q

-2q

+q

-2q

c.

39

a  0.171a

Two charges are separated by a distance 2a. Determine the location(s), if any, where the electric field is zero. Mathematical Analysis a. -2q

+3q

-2q

+4q

+4q

-2q

b.

c.

40

The plastic rod at right has charge +Q uniformly distributed along its length L. Find the electric field at all points on the x-axis to the right of the rod.

Mathematical Analysis   

Carefully identify and label “s” the location of the differential element on the diagram above. Carefully identify and label “x” the location of the point of interest on the diagram above. Write an expression for dq, the charge on the differential element.






Write an expression for



Write an integral expression for the electric field. Choosing limits of integration, calculate the integral.

rˆ , the unit vector representing the direction from the element to the point of interest.

Questions At x = L/2, what should Ex equal? Does your function agree with this observation?

If x >> L, what should Ex equal? Does your function agree with this observation?

41

The plastic rod at right has charge -Q uniformly distributed along its length L. Find the electric field at all points on the y-axis above the rod.


Carefully identify and label “s” the location of the differential element on the diagram above. Carefully identify and label “y” the location of the point of interest on the diagram above. Write an expression for dq, the charge on the differential element.











Questions For all points on the y-axis, what should Ex equal? Does your function agree with this observation?

At y = 0, what should Ey equal? Does your function agree with this observation?

If y >> L, what should Ey equal? Does your function agree with this observation?

42

The very long plastic rod at right has uniform charge density +2.0 nC/m along its length. Find the electric field 10 cm above the center of the rod.


Carefully identify and label “s” the location of the differential element on the diagram above. Carefully identify and label the location of the point of interest on the diagram above. Write an expression for dq, the charge on the differential element.











43

The 15 cm plastic rod at right has 2.5 nC distributed uniformly along its length. Find the electric field at the point (x, y) = (0, 5.0 cm).













44

The 10 cm plastic rod at right has uniform charge density –3.5 pC/m. Find the electric field at the point (x, y) = (-5.0 cm, 0).













45

The plastic rod at right extends from y = -2a to y = a and has charge Q uniformly distributed along its length. Find the electric field at the point (x, y) = (2a, 0).













46

The plastic rod at right forms a quarter-circle of radius R and has uniform charge density +. Find the electric field at the origin.


Carefully identify and label “” the location of the differential element on the diagram above. Carefully identify and label the location of the point of interest on the diagram above. Write an expression for dq, the charge on the differential element.











47

The plastic rod at right forms a half-circle and has charge -Q uniformly distributed along its length L. Find the electric field at the origin.













48

The plastic rod at right forms a three-quarter-circle of radius R and has uniform charge density +. Find the electric field at the origin.













49

The plastic rod at right has charge +Q uniformly distributed along its length L. In order to find the electric field at all points on the x-axis to the right of the rod, four different location labels are used for the differential element. For each location label, write the correct integral expression, including limits, for the electric field.

a. b. c. d. x

Mathematical Analysis a. The location of the differential element, s, is measured from the left edge of the rod.

b. The location of the differential element, s, is measured from the center of the rod.

c. The location of the differential element, s, is measured from the right edge of the rod.

d. The location of the differential element, s, is measured from the point x to the right of the rod.

50

The plastic rod at right has charge density  uniformly distributed along its length L. In order to find the electric field at all points on the y-axis above the rod, three different location labels are used for the differential element. For each location label, write the correct integral expression, including limits, for the electric field.

y a. b. c.

Mathematical Analysis a. The location of the differential element, s, is measured from the left edge of the rod.

b. The location of the differential element, s, is measured from the center of the rod.

c. The location of the differential element, s, is measured from the right edge of the rod.

51

The plastic rod at right forms a half-circle and has charge -Q uniformly distributed along its length L. In order to find the electric field at the origin, four different location labels are used for the differential element. For each location label, write the correct integral expression, including limits, for the electric field.


a. The location of the differential element, , is measured clockwise from the –y-axis.

b. The location of the differential element, , is measured clockwise from the –x-axis.

c. The location of the differential element, , is measured counterclockwise from the +y-axis.

d. The location of the differential element, , is measured counterclockwise from the +x-axis.

52

The nested spheres at right consist of an inner metal sphere of radius a and charge –Q, and an outer metal spherical shell of inner radius b, outer radius c, and charge –3Q.

Mathematical Analysis a. Find the charge density at: r = a: The charge –Q will distribute itself uniformly over the spherical surface at a. Thus, the surface charge density is given by



Q 4a 2

r = b: +Q will be drawn to the surface at r = b, so



Q 4b 2

r = c: Since the shell has a total charge of -3Q, and +Q is on the inner surface, -4Q must be on the outer surface. Therefore



 4Q 4c 2

b. Find the charge enclosed by a spherical Gaussian surface at: r < a: There is no net charge within the volume of the inner sphere since all of the excess charge is on the surface

q enclosed  0

a < r < b: The total charge on the inner sphere is within this gaussian surface

q enclosed  Q

b < r < c: This surface encloses a net charge of zero, since the charge drawn to the inner surface of the shell exactly balances the charge on the sphere.

q enclosed  0

r > c: This surface encloses all of the charge on the object

q enclosed  4Q

53

The nested spheres at right consist of an inner metal sphere of radius a and charge density +, and a neutral outer metal spherical shell of inner radius b and outer radius c.

Mathematical Analysis a. Find the total charge on the inner sphere.

b. Find the charge density at: r = b:

r = c:

c. Find the charge enclosed by a spherical Gaussian surface at: r < a:

a < r < b:

b < r < c:

r > c:

54

The nested spheres at right consist of an inner plastic sphere of radius a and charge density +, and an outer plastic spherical shell of inner radius b, outer radius c, and uniform charge density -2.

Mathematical Analysis a. Find the total charge on the inner sphere.

b. Find the total charge on the outer sphere.

c. Find the charge enclosed by a spherical Gaussian surface at: r < a:

a < r < b:

b < r < c:

r > c:

55

The coaxial cylinders at right consist of an inner metal cylinder of radius a and charge +2Q per unit length, and an outer metal cylindrical shell of inner radius b, outer radius c, and charge –3Q per unit length.

Mathematical Analysis a. Find the charge density on each of the three surfaces. r = a:

r = b:

r = c:

b. Find the charge enclosed by a cylindrical Gaussian surface of length L at: r < a:

a < r < b:

b < r < c:

r > c:

56

The coaxial cylinders at right consist of an inner plastic cylinder of radius a and uniformly distributed charge +2Q per unit length, and an outer plastic cylindrical shell of inner radius b, outer radius c, and uniformly distributed charge –3Q per unit length.

Mathematical Analysis a. Find the volume charge density, , on the inner cylinder.

b. Find the volume charge density, , on the outer cylinder.

c. Find the charge enclosed by a cylindrical Gaussian surface of length L at: r < a:

a < r < b:

b < r < c:

r > c:

57

A solid plastic sphere has radius R and charge +Q uniformly distributed throughout its volume. Find the electric field at all points in space, and sketch the radial component of the electric field on the graph at right.

Er

R

r

Mathematical Analysis for r < R:

for r > R:

  

Carefully draw the hypothetical gaussian surface at this location. Carefully draw the electric field at all points on this surface. Write an expression for the surface area parallel to the electric field.

  




Write an expression for qenclosed, the charge inside this surface.










58

A solid metal sphere has radius R and charge –Q. Find the electric field at all points in space, and sketch the radial component of the electric field on the graph at right.

Er

R

r


for r > R:

  


  














59

A long, solid plastic cylinder has radius R and uniform charge density . Find the electric field at all points in a plane perpendicular to the cylinder near its midpoint, and sketch the radial component of the electric field on the graph at right.

Er

R

r


for r > R:

  


  














60

A hollow plastic sphere has inner radius a, outer radius b, and uniform charge density . Find the electric field at all points in space, and sketch the radial component of the electric field on the graph at right.

Er

a

b

r

Mathematical Analysis for r < a: 








for a < r < b:

for r > b:



















61

A hollow metal sphere has inner radius a, outer radius b, and total charge +Q. Find the electric field at all points in space, and sketch the radial component of the electric field on the graph at right.

Er

a

b

r









for a < r < b:

for r > b:



















62

A long, hollow metal cylinder has inner radius a, outer radius b, and uniform charge density . Find the electric field at all points in a plane perpendicular to the cylinder near its midpoint, and sketch the radial component of the electric field on the graph at right.

Er

a

b

r









for a < r < b:

for r > b:



















63

A long, hollow plastic cylinder has inner radius a, outer radius b, and uniform charge density . Find the electric field at all points in a plane perpendicular to the cylinder near its midpoint, and sketch the radial component of the electric field on the graph at right.

Er

a

b

r









for a < r < b:

for r > b:



















64

A long coaxial cable has an inner wire of negligible thickness with linear charge density -and an outer conducting cylinder of inner radius a and outer radius b. The net charge on the cable is zero. Find the electric field at all points in a plane perpendicular to the cable near its midpoint, and sketch the radial component of the electric field on the graph at right.

Er

a

b

r









for a < r < b:

for r > b:



















65

A long coaxial cable has an inner wire of negligible thickness with linear charge density +and an outer conducting cylinder of inner radius a and outer radius b. The net charge on outer cylinder is zero. Find the electric field at all points in a plane perpendicular to the cable near its midpoint, and sketch the radial component of the electric field on the graph at right.

Er

a

b

r









for a < r < b:

for r > b:



















66

A large plastic plate is centered on the xz-plane. The plate has thickness 2D and charge density + uniformly distributed throughout its volume. Find the electric field at all points on the +y-axis, and sketch the y-component of the electric field on the graph at right.

Ey

D

y

Mathematical Analysis for y < D:

for y > D:

  


  














67

A large metal plate is centered on the xz plane. The plate has thickness 2D and charge density +uniformly distributed on its top and bottom surface. Find the electric field at all points on the +y-axis, and sketch the y-component of the electric field on the graph at right.

Ey

D

y

Mathematical Analysis for y < D:

for y > D:

  


  














68

Two large metal plates of negligible thickness are separated by a distance 2d and are parallel to the xz plane. The bottom plate has charge density + and the top plate –. Find the electric field at all points on the +y-axis, and sketch the y-component of the electric field on the graph at right.

Ey

d

y

Mathematical Analysis for y < d:

for y > d:

  


  














69

70