# Electromagnetic fundas

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```					Welcome…

Electric Charge
Static Electricity There are two kinds of charge. Properties of charges  like charges repel  unlike charges attract  charges can move but charge is conserved Law of conservation of charge: the net amount of electric charge produced in any process is zero. +

-

Although there are two kinds of charged particles in an atom, electrons are the charges that usually move around.

A proton is roughly 2000 times more massive than an electron. Charges are quantized (come in units of e= 1.6x10-19 C). The charge of an electron is –e = –1.6x10-19 coulombs. The charge of a proton is +e = +1.6x10-19 coulombs.

Why is the fundamental unit of charge e=1.6x10-19 C, and why is it not defined to be unity?

Nitpicking: electric charge is a property of matter, not a ―thing‖ in itself. It is ―not good‖ to say ―like charges repel.‖ It is ―good‖ to say ―like-charged particles repel.‖ I choose the ―not good‖ terminology here to be consistent with your text.

And who‘s the clown who decided electrons have negative charges? (It would be much more convenient if + charged particles were the ones that moved easily.)

And yes, he really flew the kite in the thunderstorm. See here. Franklin‘s experiments showed him that there were two ―kinds‖ of charge, which he named ―positive‖ and ―negative.‖ More than a century later we learned that negative charges are associated with electrons.
Oh, and the next two people who tried the kite experiment were killed in the process.

Induced Charge: The Electroscope
Demonstrate (if electroscope available).

Coulomb’s Law
We‘ve seen attractive and repulsive electrical forces at work. Coulomb‘s law quantifies the magnitude of the electrostatic force.
Coulomb‘s law gives the force between charges Q1 and Q2 (in coulombs), where r is the distance in meters between the charges, and k=9x109 N·m2/C2.

OSE:

 Q 1Q 2 F k , 2 12 r12

attractive for unlike

Force is a vector quantity, so the equation by itself gives the magnitude of the force. Note how the direction is specified. If the charges are opposite in sign, the force is attractive; if the charges are the same in sign, the force is repulsive. Also, the constant k is equal to 1/40, where 0=8.85x10-12 C2/N·m2. To make this into a ―really good‖ OSE I should specify ―repulsive for like,‖ but that makes it too wordy. You‘ll just have to remember that! OSE:
 Q 1Q 2 F k , 2 12 r12

attractive for unlike

Remember, a vector has a magnitude and a direction.

The equation is valid for point charges. If the charged objects are spherical and the charge is uniformly distributed, r is the distance between the centers of the spheres. r +

-

If more than one charge is involved, the net force is the vector sum of all forces (superposition—let‘s not make that a required OSE). For objects with complex shapes, you must add up all the forces acting on each separate charge (turns into calculus!). + + + -

We could have agreed that in the formula for F, the symbols Q1 and Q2 stand for the magnitudes of the charges. In that case, the absolute value signs would be unnecessary. However, in later OSE‘s the sign of the charge will be important, so we really need to keep the magnitude part. On your homework diagrams, show both the magnitudes and signs of Q1 and Q2. Your starting equation sheet has this version of the equation:
F k 12 Q 1Q 2 r
2 12

,

which is giving you the magnitude and implying that you need to figure out the direction separately.

A sample Coulomb‘s law problem involving multiple charges is on the following slides.

I will work the problem on the blackboard in lecture, if there is time. Usually vector problems are easiest if you manipulate the ―whole‖ vector at once, using unit vectors. Sometimes it is easier to work the problem a component at a time. The slides use the component-at-a-time approach. The work in class will use the unit vector approach. I recommend the unit vector approach.

Solving Problems Involving Coulomb’s Law and Vectors
You may wish to review vectors.

Example: Calculate the net electrostatic force on charge Q3 due to the charges Q1 and Q2.
y
Q3=+65C
30 cm

=30º Q2=+50C 52 cm Q1=-86C

x

Step 0: Think!
This is a Coulomb‘s Law problem (all we have to work with, so far). We only want the forces on Q3. Don‘t worry about other forces. Forces are additive, so we can calculate F32 and F31 and add the two.
If we do our vector addition using components, we must resolve our forces into their x- and y-components.

Step 1: Diagram
F32 Draw a representative sketch—done. Q3=+65C
30 cm

y

Draw and label relevant quantities—done.

F31
=30º Q1=-86C
52 cm

Draw axes, showing origin and directions— Q2=+50C done. Draw and label forces (only those on Q3).

x

Draw components of forces which are not along axes.

Step 2: OSE
y
Q3=+65C 30 cm

F32 F31
=30º

Q2=+50C
52 cm

Q1=-86C

x

 Q 1Q 2 F k , attractive for unlike 2 12 r12

<complaining> ”Do I have to put in the absolute value signs?” Yes.

Step 3: Replace Generic Quantities by Specifics
y
 Q 3Q 2 F  k , 2 32 r3 2 re p u lsive
Q3=+65C 30 cm

F32 F31
=30º

F k 32, y

Q 3Q 2 r
2 32

Q2=+50C
52 cm

Q1=-86C

x

F  0 32, x

(from diagram)

Can you put numbers in at this point? OK for this problem. You would get F32,y = 330 N and F32,x = 0 N.

Step 3 (continued)
 Q 3Q 1 F k , 2 31 r3 1 a ttra ctive

y
Q3=+65C 30 cm

F32 F31
=30º

F  k 31, x

Q 3Q 1 r
2 31

cos 
Q2=+50C

(+ sign comes from diagram)
F  k 31, y Q 3Q 1 r
2 31

Q1=-86C
52 cm

x

sin 

(- sign comes from diagram)

Can you put numbers in at this point? OK for this problem. You would get F31,x = +120 N and F31,y = -70 N.

Step 3: Complete the Math
y
The net force is the vector sum of all the forces on Q3.
Q3=+65C 30 cm

F32 F31

F3

=30º Q2=+50C
52 cm

Q1=-86C

x

F3x = F31,x + F32,x = 120 N + 0 N = 120 N
F3y = F31,y + F32,y = -70 N + 330 N = 260 N

You know how to calculate the magnitude F3 and the angle between F3 and the x-axis. (If not, holler!)

Remember a better way to write Coulomb‘s law than the following? OSE:
 Q 1Q 2 F k , attractive for unlike 2 12 r12

r If you define ˆ1  2 as the unit vector (a vector of unit length) pointing from Q2 to Q1, and F12 is the force on Q1 due to Q2, and you include the correct signs in Q1 and Q2, then

 Q 1Q 2 ˆ1 2 F  k r 2 12 r1 2

+ Q1 r12 Q2

Much more satisfying!

Coulomb’s Law Applied to Charge Distributions
In Monday‘s lecture you were introduced to Coulomb‘s Law, which lets you calculate the electrostatic force between two charged point particles or spherically-symmetric charge distributions:
q 1q 2 r
2 12

r12
,

F =k 12

+ Q1

Q2

k = 8.988x109 Nm2/C2 (and it is usually OK to round k to 9x109.

F = 12 4πε 0 1 q 1q 2 r
2 12

,

where 0 = 8.854x10-12 C2/Nm2. Both expressions of Coulomb‘s Law are valid. You might want to get used to using the one with 0, because that constant will be with us the rest of the semester, in a variety of other equations. By itself, this is not particularly useful, but if we could apply it to ―real‖ systems, we might be able to do useful calculations.

It is relatively easy to calculate the force on one charged particle due to a few others.*
 F1 =


i

 F1 i =

q1 4 πε 0


i

qi  r 1 i . 2 r1 i

This is not a separate OSE; it is simply an application of superposition to Coulomb‘s Law.

Would you like to try that for ―many‖ (1023) charged particles?

*Avoid this common mistake: F1+F2=F1+F2.

In today's lecture I show you how to extend Coulomb's law to the case of a single point (or spherically symmetric) charge interacting with a distribution of charges. Matter is made of discrete atoms, but appears "continuous" to us, and in Physics 23 we treated matter as being a continuous entity. Similarly, a charge distribution is made of individual charged particles, but we can treat it as if the charge were continuous.

The force on a point charge q due to a small "chunk" q of charge is
 F = qq  r . 2 4 πε 0 r 1

unit vector from q to q The force on a point charge q due to collection of "chunks" of charge is
 F =


i

  Fi =

q 4 πε 0


i

qi  ri. 2 ri

unit vector from qi to q

If charge is distributed along a straight line segment parallel to the x-axis, the amount of charge dq on a segment of length dx is dx.  dx
x dx  is the linear density of charge (amount of charge per unit length).  may be a function of position.

Think     length.  times the length of line segment is the total charge on the line segment.

q r‘
 r'

x

dq The force on a point charge q due to the charge dq is
 dF = q dq  1 q dx  r' = r'. 2 2 4 πε 0 r' 4 πε 0 r' 1

q r‘
 r'

x

dq The net force on q due to the entire line of charge is
 F=  q λ(x ) d x = q  r' r'2 4 πε 0 4 πε 0 1  λ(x ) d x .  r' r'2

The integration is carried out over the entire length of the line, which need not be straight. Also,  could be a function of position, and can be taken outside the integral only if the charge distribution is uniform.

If charge is distributed over a two-dimensional surface, the amount of charge dq on an infinitesimal piece of the surface is dS, where  is the surface density of charge (amount of charge per unit area).
y



charge dq =  dS

x

area = dS

y q r‘
 r'

x

The force on a point charge q due to the charge dq is
 dF = q dq  1 q  dS  r' = r'. 2 2 4 πε 0 r' 4πε 0 r' 1

y q r‘
 r'

x

The net force on q due to the entire surface of charge is
 F=  q  (x, y) dS = q 2  r' 4πε 0 S r' 4πε 0 1   (x, y) dS .  r' r' 2 S

After you have seen the above the above, I hope believe that the net force on a charge q due to a three-dimensional distribution of charge is z q

r‘
 r'

x

y
 F=  q  (x, y, z) dV = q 2  r' 4πε 0 V r' 4πε 0 1   (x, y, z) dV . 2  r' r' V

Summarizing: Charge distributed along a line:

 F=

 λ dx .  r' r'2 4 πε 0 q

Charge distributed over a surface:

 F=

  dS .  r' r'2 4πε 0 S q   dV .  r' r'2 4πε 0 V q

Charge distributed inside a volume:

 F=

If the charge distribution is uniform, then , , and  can be taken outside the integrals.

Example (worked at blackboard): calculate the net force on a point charge q placed at the center of a uniformly charged ring.

Example (worked at blackboard): calculate the net force on a point charge Q placed on the axis of a ring of charge a distance D away from the center of the ring. These examples are worked in your text.

The Electric Field
Coulomb's Law (demonstrated in 1785) shows that charged particles exert forces on each other over great distances. How does a charged particle "know" another one is there?

Action At A Distance Viewpoint
Electric, magnetic, and gravitational forces result of direct and instantaneous interaction between particles.

Faraday showed why this viewpoint is wrong.

Faraday (not much at math, but great in the lab), beginning in 1830's, was the leader in developing the idea of the electric field. Here's the idea:
 A charged particle emanates a "field" into all space.  Another charged particle senses the field, and ―knows‖ that the first one is there. The idea of an electric field is good for a number of reasons:

It makes us feel good, like we‘ve explained something.
We can develop a theory based on this idea. From this theory may spring unimagined inventions.

We start by defining an electric field by the force it exerts on a test charge q0:
 E=  F q0
This is your second starting equation. By convention the direction of the electric field is the direction of the force exerted on a POSITIVE test charge.

If the test charge is "too big" it perturbs the electric field, so the ―correct‖ definition is
 E = lim  F q0

q0  0

Let‘s see what our ‗baby‘ can do.

The Electric Field Due to a Point Charge
Coulomb's law says
F =k 12 q 1q 2 r
2 12

,

... which tells us the electric field due to a point charge is
 q ˆ E q =k 2 r r
This is your third starting equation.

The electric field direction is the direction a + charge would feel a force. A + charge would be repelled by another + charge. Therefore the direction of the electric field is away from positive (and towards negative).

 q E q =k 2 , aw ay from + r

http://regentsprep.org/Regents/physics/phys03/afieldint/default.htm

Example: calculate the electric field at the electron‘s distance away from the proton in a hydrogen atom (5.3x10-11 m).

The Electric Field Due to a Collection of Point Charges
 q ˆ 2i E n et = k  r ri i

Example: calculate the electric field at a point P which lies on the perpendicular bisector of a dipole.

Electric Field Lines
Electric field lines help us visualize the electric field and predict how charged particles would respond the field.

Example: electric field lines for isolated +2e and -e charges.

Here‘s how electric field lines are related to the field:

 The electric field vector E is tangent to the field lines.  The number of lines per unit area through a surface perpendicular to the lines is proportional to the electric field strength in that region  The field lines begin on positive charges and end on negative charges.  The number of lines leaving a positive charge or approaching a negative charge is proportional to the magnitude of the charge.  No two field lines can cross.
Example: draw the electric field lines for charges +2e and -1e, separated by a fixed distance.

Whilst attempting to explain a discovery to either Lord Gladstone, Chancellor of the Exchequer, Faraday was asked, ―But, after all, what use is it?‖ Faraday replied, ―Why sir, there is the probability that you will soon be able to tax it.‖

The Electric Field Due to a Continuous Charge Distribution

dq r

dE

 dq ˆ dE = k 2 r r

 E = lim

dq  0



 dq ˆ 2 dE =k  r r

Example: A rod of length L has a uniform charge per unit length  and a total charge Q. Calculate the electric field at a point P along the axis of the rod at a distance d from one end.
y

P d L

x

Let‘s put the origin at P. The linear charge density and Q are related by
 = Q L and Q =  L

Let‘s assume Q is positive.

y dE
P

x d

dx L

dQ =  dx

x

The electric field points away from the rod. By symmetry, the electric field on the axis of the rod has no y-component. dE from the charge on an infinitesimal length dx of rod is
dE = k dq x
2

k

 dx x
2

y dE
P

x d

dx L

dQ =  dx

x

E = k

d+L d

 dx x
2

= k 

d+L d

dx x
2

 1 = k     x d

dL

 d  d  L  1 1  L kQ  E = k     = k  = k =  d  d  L   d  d  L d  d  L  dL d  

Confession
I have a Confession to make.
 q E q =k 2 , aw ay from + r

is (at best) possibly misleading. If you include the sign on q, you may end up with E pointing in the wrong direction.

charge magnitude = Q E x = -k q r
2

Ex
(-Q ) r
2

x
kQ r
2

= -k

=

Correct and unambiguous:

 q E q =k 2 , aw ay from + r

The Electric Field Due to a Continuous Charge Distribution (conclusion)
Example: A ring of radius a has a uniform charge per unit length and a total positive charge Q. Calculate the electric field at a point P along the axis of the ring at a distance x0 from its center.
dQ a x0
r



P 

x

dE

By symmetry, the y- and zcomponents of E are zero, and all points on the ring are a distance r from point P.

dQ

dE = k
a
x0

dQ r
2

r
 P


dE

x

dE x =k

dQ r
2

cos 

r=

x0  a
2

2

cos  

x0 r

For a given x0, r is a constant for points on the ring.

x0  dQ  x 0 E x   dE x    k 2  k 3 r  r r ring ring 


ring

dQ  k

x0 r
3

Q 

kx 0 Q

x
.

2 0

a

2



3/2

Or, in general, on the ring axis E x , ring 

kxQ

x  a
2

2



3/2

Example: A disc of radius R has a uniform charge per unit area . Calculate the electric field at a point P along the central axis of the disc at a distance x0 from its center.
dQ
r

P
R

x

x0

The disc is made of concentric rings. The area of a ring at a radius r is 2rdr, and the charge on each ring is (2rdr).

We can use the equation on the previous slide for the electric field due to a ring, replace a by r, and integrate from r=0 to r=R.
dE ring  kx 0  2  rdr
2 2

x

r

2



3/2

.

dQ r

P
R

x

x0

Ex 


disc

dE x 


disc

kx 0  2  rdr

x

2 0

r

2



3/2

 kx 0  

R 0

2 r dr

x

2 0

r

2



3/2

 x2  r2    0 E x  kx 0   1 / 2 

1 / 2

  x x0   2 k   0  1/ 2 2 2  x0  x0  R  0 

R

   

Example: Calculate the electric field at a distance x0 from an infinite plane sheet with a uniform charge density . Treat the infinite sheet as disc of infinite radius. Let R and use k 
1 4  0

to get
 20

E x ,sh eet 

.

Interesting...does not depend on distance from the sheet.

Motion of a Charged Particle in a Uniform Electric Field
A charged particle in an electric field experiences a force, and if it is free to move, an acceleration. If the only force is due to the electric field, then
- - - - - - - - - - - - -



   F  m a  qE.

F

E

+ + + + + + + + + + + + +

If E is constant, than a is constant, and you can use the equations of kinematics* (remember way back to the beginning of physics 23?).
*If you get called to the board, you can use the Physics 23 starting equations. They are posted.

Example: a proton and an electron enter a region of uniform electric field. Describe their motion. Direction of forces? Magnitudes of accelerations? Shape of trajectories?

Example: an electron moving with velocity v0 in the positive x direction enters a region of uniform electric field that makes a right angle with the electron‘s initial velocity. Express the position and velocity of the electron as a function of time.
y
- - - - - - - - - - - - -

x
v0

E

+ + + + + + + + + + + + +

Demo: Cathode Ray Tube

Demo: Professor Tries to Avoid Shocking Himself While Demonstrating Van de Graaff Accelerator

Electric Dipole in an External Electric Field
An electric dipole consists of two charges +q and -q, equal in magnitude but opposite in sign, separated by a fixed distance L. q is the ―charge on the dipole.‖ In the previous lecture, I calculated the electric field along the perpendicular bisector of a dipole.
E  qL 4  o r
3

.

The electric field depends on the product qL.

q and L are parameters that specify the dipole; we define the "dipole moment" of a dipole to be the vector
  p  qL ,

where the direction of p is from negative to positive (NOT away from +).
+q -q

L

A dipole in a uniform electric field experiences no net force, but probably experiences a torque.

Noooooooo! No torques!

There are worse things on this earth than torques…

from http://www.nearingzero.net (nz262.jpg)

A dipole in a uniform electric field experiences no net force, but probably experiences a torque… F+ E

p F-q


+q

There is no net force on the dipole:



     F  F  F   qE  qE  0.

p ½ L sin F-q 

+q

F+

E
½ L sin

If we choose the midpoint of the dipole as the origin for calculating the torque, we find





  

L sin  2

qE 

L sin  2

qE  qLE sin  ,

and in this case the direction is into the plane of the figure. Expressed as a vector,
     p  E.

p ½ L sin F-q 

+q

F+

E
½ L sin

The torque‘s magnitude is pEsin and the direction is given by the right-hand rule.

Energy of an Electric Dipole in an External Electric Field
p
F-q  +q

F+

E

If the dipole is free to rotate, the electric field does work to rotate the dipole.
W   pE (cos  initial  cos  final ).

The work depends only on the initial and final coordinates, and not on how you go from initial to final.

Does that awaken vague memories of Physics 23?
If a force is conservative, you can define a potential energy associated with it.

What kinds of potential energies did you learn about in Physics 23?
Because the electric force is conservative, we can define a potential energy for a dipole. The equation for work
W   pE (cos  initial  cos  final )

tells us we need to define
U d ip o le   p E co s  .

U dipole   pE cos 

p F-q


+q

F+

E

U is zero when =/2 (makes sense, see picture). U is maximum when cos=-1, or = (a point of unstable equilibrium).
U is minimum when cos=+1, or =0 (stable equilibrium).

You can also express the dipole potential energy as
U dipole     p  E.

Gauss’ Law
Electric Flux
We have used electric field lines to visualize electric fields and indicate their strength. We are now going to count the number of electric field lines passing through a surface, and use this count to determine the electric field.

E

The electric flux passing through a surface is the number of electric field lines that pass through it.

Because electric field lines are drawn arbitrarily, we quantify electric flux like this: E=EA.

A

E

If the surface is tilted, fewer lines cut the surface.
Later we‘ll learn about magnetic flux, which is why I will use the subscript E on electric flux.

E 

We define A to be a vector having a magnitude equal to the area of the surface, in a direction normal to the surface.

 E 

A

The ―amount of surface‖ perpendicular to the electric field is Acos . Because A is perpendicular to the surface, the amount of A parallel to the electric field is Acos .
A = A cos  so E = EA = EA cos .

Remember the dot product from Physics 23?  E

   E A

If the electric field is not uniform, or the surface is not flat…

divide the surface into infinitesimal surface elements and add the flux through each… E
 E  lim
A i  0

dA


i

  E i  A i

E

    E  dA

If the surface is closed (completely encloses a volume)…

…we count lines going out as positive and lines going in as negative… E dA
E     E  dA 

a surface integral, therefore a double integral 

What the *!@* is this

 

thing?

The circle reminds you to integrate over a closed surface.

Example: Calculate the electric flux through a cylinder with an axis parallel to the electric field direction.

E

To be worked at the blackboard…

E

If there were a + charge inside the cylinder, there would be more lines going out than in. If there were a - charge inside the cylinder, there would be more lines going in than out.

Gauss’ Law
Mathematically, we express the idea on the previous slide as
E   q en clo sed   E  dA   o

Gauss‘ Law

We will find that Gauss law gives a simple way to calculate electric fields for charge distributions that exhibit a high degree of symmetry…

…and save more complex and realistic charge distributions for advanced classes.

―Mathematics is the Queen of the Sciences.‖—Karl Gauss

To see how this works, let‘s do an example. Example: use Gauss‘ Law to calculate the electric field from an isolated point charge q.
To apply Gauss‘ Law, we construct a ―Gaussian Surface‖ enclosing the charge.

The Gaussian surface should mimic the symmetry of the charge distribution. For this example, choose for our Gaussian surface a sphere of radius r, with the point charge at the center.
I‘ll work the rest of the example on the blackboard.

Strategy for Solving Gauss’ Law Problems
 Select a Gaussian surface with symmetry that matches the charge distribution.

 Draw the Gaussian surface so that the electric field is either constant or zero at all points on the Gaussian surface.  Use symmetry to determine the direction of E on the Gaussian surface.
 Evaluate the surface integral (electric flux).

 Determine the charge inside the Gaussian surface.
 Solve for E.
If I were Dr. Bieniek, this would be a litany.

Example: calculate the electric field outside a long cylinder of finite radius R with a uniform volume charge density  spread throughout the volume of the cylinder. The cylinder being ―long‖ and the radius ―finite‖ are ―code words‖ that tell you to neglect end effects from the cylinder (i.e., assume it is infinitely long). Know how to interpret ―code words‖ when exam time comes! Let‘s go through this a step at a time on the blackboard.
 Select a Gaussian surface with symmetry that matches the charge distribution.

Pick a cylinder of length L and radius r, concentric with the cylinder of the problem.

 Draw the Gaussian surface so that the electric field is either constant or zero at all points on the Gaussian surface. Already done!
 Use symmetry to determine the direction of E on the Gaussian surface. Electric field points radially away from cylinder, and magnitude does not depend on direction.

 Evaluate the surface integral (electric flux). Surface integral is just E times the curved area.

 Determine the charge inside the Gaussian surface. The charge inside is just the volume of charged cylinder inside, times the charge per volume.

 Solve for E.
E R
2

2 0r

More examples at end of lecture, if time permits.

Conductors in Electrostatic Equilibrium
Electrostatic equilibrium means there is no net motion of tne charges inside the conductor. The electric field inside must be zero. If this were not the case, charges would move. Any excess charge must reside on the outside surface of the conductor. Apply Gauss‘ law to a Gaussian surface just inside the conductor surface. The electric field is zero, so the net charge inside the Gaussian surface is zero.

The electric field just outside a charged conductor must be parallel to the conductor‘s surface. Otherwise, the component of the electric field parallel to the surface would cause charges to accelerate. The magnitude of the electric field just outside a charged conductor is equal to /0, where  is the local surface charge density.
A simple application Gauss‘ Law, which I will show if time permits.

If there is an empty nonconducting cavity inside the conductor, Gauss‘ Law tells us there is no net charge on the interior surface of the conductor. If there is a nonconducting cavity inside the conductor, with a net charge inside the cavity, Gauss‘ Law tells us there is an equal and opposite induced charge on the interior surface of the conductor.

Example: a conducting spherical shell of inner radius a and outer radius b with a net charge -Q is centered on point charge +2Q. Use Gauss‘s law to show that there is a net charge of -2Q on the inner surface of the shell, and a net charge of +Q on the outer surface of the shell.

-Q

a
+2Q b

  q en clo sed  E  dA    o

Example (if time permits): an insulating sphere of radius a has a uniform charge density ρ and a total positive charge Q. Calculate the electric field at a point inside the sphere

r

a

  q en clo sed  E  dA    o

Q

Electric Potential Electric Potential Energy
Electric Potential Energy
Work done by Coulomb force when q1 moves from a to b:
W 

r



rb ra

 rb k q q  1 2 FE  d s    dr 2 ra r

dr
rb

b q1 (+)

ds rab

W   k q 1q 2



rb ra

1 r
2

 1 dr   k q 1 q 2     r

ra

a ra

FE

 1  1 1  1  W   k q 1q 2      k q 1q 2     rb ra   rb ra 

q2 (-)

 1 1  W  k q 1q 2     rb ra 

r

dr FE

b q1 (+)

ds rab

I did the calculation for a + charge moving away from a – charge; you could do a similar calculation for ++, -+, and ++.
The important point is that the work depends only on the initial and final positions of q1.

a ra

q2 (-)

In other words, the work done by the electric force is independent of path taken. The electric force is a conservative force.
Disclaimer: this is a ―demonstration‖ rather than a rigorous proof.

A bit of review: Consider an object of mass m in a gravitational field. It has potential energy U(y) = mgy and ―feels‖ a gravitational force FG = GmM/r2, attractive.
y

If released, it gains kinetic energy and loses potential energy, but mechanical energy is conserved: Ef=Ei. The change in potential energy is Uf - Ui = -(Wc)if.
What force does Wc? Force due to gravity.

Ui = mgyi yi
x Uf = 0

graphic ―borrowed‖ from http://csep10.phys.utk.edu/astr161/lect/history/newtongrav.html

A charged particle in an electric field has electric potential energy. It ―feels‖ a force (as given by Coulomb‘s law). It gains kinetic energy and loses potential energy if released, and its mechanical energy is conserved.

++++++++++++++

+ E

-------------------

Now that we realize the electric force is conservative, we can define a potential energy associated with it.
 U  U f  U i    W E i  f

The potential energy change when a charge q0 moves from point a to point b in the electric field of charge q is
U   
rb ra

r
dr

b q0

 rb k qq  0 FE  d s    dr 2 ra r

ds
rab

The minus sign in this equation comes from the definition of change in potential energy. The sign from the dot product is ―automatically‖ included if you include the signs of q and q0.

a ra

FE ?

q

? on FE means the direction depends on the signs of the charges.

U b  U a   k qq 0 

rb ra

1 r
2

 1 dr   k qq 0     r

rb

r
dr

b q0

ra

ds
rab

 1 1  U b  U a  k qq 0     rb ra 

By convention, we choose electric potential energy to be zero at infinite separation of the charges.

a ra

FE ?

q

If there are any math majors in the room, please close your eyes for a few seconds. 0
Ub  U

0  1 1   k qq 0     rb  

This provides us with the electric potential energy for a system of two point charges q and q0, separated by a distance r:
U  r   k qq 0 1 r  qq 0 1 4  0 r

.

Example: calculate the electric potential energy for two protons separated by a typical proton-proton internuclear distance of 2x10-15 m. What is the meaning of the + sign in the result?

E f  E i   W other i  f

What ―goes into‖ Ef and Ei? What ―goes into‖ Wother?

Electric Potential
In lecture 3 we defined the electric field by the force it exerts on a test charge q0: 
 E = lim F q0
q0  0

It is useful to define the potential of a charge (or charge distribution) in terms of the potential energy of a test charge q0:
 V  r  = lim  Ur q0
q0  0

The electric potential V is independent of the test charge q0.

From U  r  

qq 0 1 4  0 r

we see that the electric potential of a point charge q is
V r  q 1 4  0 r .

The electric potential difference between points a and b is
V  U q0  
rb ra

  FE  d s q0

 

rb ra

 FE q0

rb    ds   E ds . ra

Things to remember about electric potential:
 Electric potential and electric potential energy are related, but not the same.

Electric potential difference is the work per unit of charge that must be done to move a charge from one point to another without changing its kinetic energy.  The terms ―electric potential‖ and ―potential‖ are used interchangeably.
  The units of potential are joules/coulomb: V  r  =
1 volt = 1 joule 1 coulom b

 Ur q0

.

Things to remember about electric potential:  Only differences in electric potential and electric potential energy are meaningful. It is always necessary to define where U and V are zero. We have defined V to be zero at an infinite distance from the sources of the electric field. Sometimes it is convenient to define V to be zero at the earth (ground).
It should be clear from the context where V is defined to be zero, and I do not foresee you experiencing any confusion about where V is zero.

This lecture has been rather abstract so far. Let‘s do two conceptual examples. Example: a proton is released in a region in space where there is an electric potential. Describe the subsequent motion of the proton.
The proton will move towards the region of lower potential. As it moves, its potential energy will decrease, and its kinetic energy and speed will increase.

Example: an electron is released in a region in space where there is an electric potential. Describe the subsequent motion of the electron.
The electron will move towards the region of higher potential. As it moves, its potential energy will decrease, and its kinetic energy and speed will increase.

Electric Potential Energy of a System of Charges To find the electric potential energy for a system of two charges, we bring a second charge in from an infinite distance away:
r q1 q1 q2
q 1q 2 r

U 0

U  k

before

after

To find the electric potential energy for a system of three charges, we bring a third charge in from an infinite distance away: r12 q1 q2
q1

r12
q2

r13 q3

r23

before
U  k q 1q 2 r1 2

after
 q 1q 2 q 1q 3 q 2 q 3  U  k    r13 r23   r12

Electric Potential and Potential Energy of a Charge Distribution Collection of charges: V P 
1 4  0


i

qi ri

.

P is the point at which V is to be calculated, and ri is the distance of the ith charge from P.

Charge distribution:
VP  1 4  0

dq



dq r

.

P

r

Example: a 1 C point charge is located at the origin and a -4 C point charge 4 meters along the +x axis. Calculate the electric potential at a point P, 3 meters along the +y axis.

y
VP = k 
i

qi

P 3m
q2 x

 q1 q 2  = k +  ri r1 r2  

 1×10 -6 -4 ×10 -6  9 = 9 ×10  +  3 5   = - 4.2 ×10 V
3

q1

4m

Thanks to Dr. Waddill for the use of these examples.

Example: how much work is required to bring a +3 C point charge from infinity to point P?
y

0
W external   E   K   U W external   U  q 3  V

q3 P
3m q2 x

W external  q 3  VP  V 
4m

0
3

q1

W external  3  10

6

  4.2  10
3



W external   1.26  10 J
The work done by an external force is negative, and the work done by the electric field was positive. The electric field ―pulled‖ q3 in (keep in mind q2 is 4 times as big as q1). Positive work would have to be done by an external force to remove q3 from P.

Example: find the total potential energy of the system of three charges.
y

q3 P
3m q2 x

 q1 q 2 q1 q 3 q 2 q 3  U = k + +  r12 r13 r23  

q1

4m

-6 -6 -6 -6  1×10 -6 -4 ×10 -6 1×10 3 ×10 -4 ×10 3 ×10 9 U = 9  10  + +  4 3 5 





 



 



 
 

U = - 2.16  10

-2

J

The Electron Volt An electron volt (eV) is the energy acquired by a particle of charge e when it moves through a potential difference of 1 volt.
U= qV
1 eV = 1.6  10
-19

C  1 V 
-19

1 eV= 1.6  10

J

This is a very small amount of energy on a macroscopic scale, but electrons in atoms typically have a few eV (10‘s to 1000‘s) of energy.

Equipotentials
Equipotentials are contour maps of the electric potential.

http://www.omnimap.com/catalog/digital/topo.htm

Equipotential lines are another visualization tool. They illustrate where the potential is constant. Equipotential lines are actually projections on a 2-dimensional page of a 3dimensional equipotential surface. (―Just like‖ the contour map.) The electric field must be perpendicular to equipotential lines. Why?
Otherwise work would be required to move a charge along an equipotential surface, and it would not be equipotential.

In the static case (charges not moving) the surface of a conductor is an equipotential surface. Why?
Otherwise charge would flow and it wouldn‘t be a static case.

Here are some electric field and equipotential lines I generated using the an electromagnetic field program.

Equipotential lines are shown in red.

 Starting Equation Confusion:
 U  U b  U a   W ba

Wba is the work done by the conservative force that gives rise to the potential, and should really be written Wab.

A less ambiguous version of the equation would be
 U  U f  U i    W conservative i  f .

If you need to calculate the work done by an external force, you really should start with
E f  E i   E   W other i  f .

 More Confusion:
Fk q 1q 2 r
2

 q 1q 2 ˆ Fk 2 r r
No absolute value signs mean you must use the signed values of charge and sign convention for r.

Absolute value signs mean you have to figure out direction of F for yourself.

V k

q r

U k

q 1q 2 r1 2

No absolute value signs mean you must use the signed values of charge.

r is a distance, so it must be positive. Should I put absolute value signs around the r‘s?

 Speed and kinetic energy:
E f  E i   E   W other i  f .

This work-energy equation tells you how to calculate kinetic energies and speeds of charged particles moving in electric fields. You haven‘t had any homework problems asking you to calculate speeds. Tomorrow you‘ll get a practice problem asking you to calculate a speed. Make sure you understand how to do it.

Equipotentials
Equipotentials are contour maps of the electric potential.

http://www.omnimap.com/catalog/digital/topo.htm

Equipotential lines are another visualization tool. They illustrate where the potential is constant. Equipotential lines are actually projections on a 2-dimensional page of a 3dimensional equipotential surface. (―Just like‖ the contour map.) The electric field must be perpendicular to equipotential lines. Why?
Otherwise work would be required to move a charge along an equipotential surface, and it would not be equipotential.

In the static case (charges not moving) the surface of a conductor is an equipotential surface. Why?
Otherwise charge would flow and it wouldn‘t be a static case.

Here are some electric field and equipotential lines I generated using an electromagnetic field program.

Equipotential lines are shown in red.

Determining Electric Field from Potential
The electric field vector points from higher to lower potentials.

More specifically, E points along shortest distance from a higher equipotential surface to a lower equipotential surface. You can use E to calculate V:
Vb  Va   
b a

  E ds .

You can use the differential version of this equation to calculate E from a known V:
  dV   E  d s   E s ds  Es   dV ds

For spherically symmetric charge distribution:
Er   dV dr

In one dimension:
Ex   dV dx

In three dimensions:
Ex   V x , Ey   V y , Ez   V z .

Example: in homework problem 24.26, you calculated a potential of the form
V (x) = A x B C-x ,

where A, B, and C were constants. What is the x-component of the electric field along the line where the equation for V is valid?
  B A B  A      2     2  Ex     1  2  2  x x x   C  x    C  x     V

This is valid only along the line between the two charges, where the equation for V(x) is valid. This technique gives you the magnitude E, and you may have to use your brain to determine the direction of E.

Potentials and Fields Near Conductors
When there is a net flow of charge inside a conductor, the physics is generally complex.

When there is no net flow of charge, or no flow at all (the electrostatic case), then a number of conclusions can be reached using Gauss‘ Law and the concepts of electric fields and potentials…

Summary of key points (electrostatic case): The electric field inside a conductor is zero. Any net charge on the conductor lies on the outer surface. The potential on the surface of a conductor, and everywhere inside, is the same. The electric field just outside a conductor must be perpendicular to the surface. Equipotential surfaces just outside the conductor must be parallel to the conductor‘s surface.

Another key point: the charge density on a conductor surface will vary if the surface is irregular, and surface charge collects at ―sharp points.‖ Therefore the electric field is large (and can be huge) near ―sharp points.‖

Another Practical Application
To best shock somebody, don‘t touch them with your hand; touch them with your fingertip. Better yet, hold a small piece of bare wire in your hand and gently touch them with that.

How to Calculate the Potential of a Charge Distribution
Example: potential and electric field between two parallel conducting plates. Assume V0<V1 (so we can determine the direction of the electric field. Also assume the plates are large compared to their separation, so the electric field is constant and perpendicular to the plates.
Also, let the plates be separated by a distance L. E

V0

L

V1

 V  V1  V 0   

plate 1 plate 0

  E ds
L

y x

V   

L 0

E

dx   E  dx  E L
0

z

E ds L V1

E 

V L

V0

Example: A rod of length L located along the x-axis has a uniform charge per unit length  and a total charge Q. Find the electric potential at a point P along the y-axis a distance d from the origin.
y
P

=Q/L
r

d

dq x
L

dq=dx
x

dx

dV  k

dq r

k

dx x d
2 2

V 



L

dV
0

Thanks to Dr. Waddill for this fine example.

V 
y P



L

k
0

 dx x d
2 2

k

Q L



L 0

dx x d
2 2

d

r

dq x
L

A good set of math tables will have the integral:
x

dx



dx x d
2 2

 ln x 



x d
2

2



L V  ln   L  kQ

2 2 L d    d 

Example: Find the electric potential due to a uniformly charged ring of radius R and total charge Q at a point P on the axis of the ring.
dQ R
x r P x

Every dQ of charge on the ring is the same distance from the point P.

dV  k

dq r

k

dq x R
2 2

V 



rin g

dV  k 

dq
rin g

x R
2

2

dQ R
x r P x

V 

k x R
2 2



dq
rin g

V 

kQ x R
2 2

Could you use this expression for V to calculate E? Would you get the same result as I got in Lecture 4?

Example: A disc of radius R has a uniform charge per unit area  and total charge Q. Calculate V at a point P along the central axis of the disc at a distance x from its center.
dQ
r

P
R

x

x

The disc is made of concentric rings. The area of a ring at a radius r is 2rdr, and the charge on each ring is (2rdr).

We can use the equation on the previous slide for the potential due to a ring, replace R by r, and integrate from r=0 to r=R.
d V rin g  k  2  rdr x r
2 2

dQ r

P R
x

x

V 


R

dV 
rin g

1 4  0



 2  rdr
rin g

x r
2



 2 0

2



R 0

rdr x r
2 2

V 

 20

x r
2

2 0



 20



x R
2

2

x 



Q 2  0 R
2


2

x R
2

2

x





Q R

dQ r

P R
x

x

V 

Q 2  0 R
2



x R
2

2

x



Could you use this expression for V to calculate E? Would you get the same result as I got in Lecture 4?

Review

Three charges +Q, +Q, and –Q, are located at the corners of an equilateral triangle with sides of length a. What is the force on the charge located at point P (see diagram)?
y P +Q

F1
 

 F   F1 cos   F2 cos   ˆ i   F1 sin   F2 sin   ˆ j
2   kQ 2 kQ   ˆ F   2 cos 60  2 cos 60  i a  a  2  kQ 2 kQ   ˆ   2 sin 60  2 sin 60  j a  a 

F2 a

+Q

-Q

x

y P +Q

F1
 

2  kQ  F  2 2 cos 60 ˆ i a

F2 a

 kQ 2 ˆ F i 2 a

+Q

-Q

x

What is the electric field at P due to the two charges at the base of the triangle? You can repeat the above calculation, replacing F by E.
 Or… E   F q
2 kQ a
2 2  cos 60 ˆ i

y P +Q

F1
 

F2 a

 E 

Q

+Q

-Q

x

 kQ  E  2 2 cos 60 ˆ i a

A thin rod of length L with total charge Q lies along the x-axis as shown. Find the magnitude and direction of the electric field at P, a distance y from the rod and along the perpendicular bisector of the rod. An infinitesimal length dx of rod has dx of charge, where =Q/L.
y P y x

Ex from a dq of charge with x=|x1|>0 will be equal and magnitude and opposite in direction to Ex from a dq of charge with x=-|x1|<0.

L

Thus, by symmetry, Ex,net=0.

dE y  dE cos  
x y

kdq r
2

cos  

k  dx r
2

cos 

tan  

cos  

y r

dx  y d  tan   

y cos 
2

d

dE  y r
P 

dx r
2



y r cos 
2 2

d

y

dx
dq L
x

r

2



y y r    r 
2 2

d 

1 y

d

dE y  dE cos  

kdq r
2

cos  

k  dx r
2

cos  

kd y

cos 

Limits of integration are from 0 to 1, where
dE  y r
P 

y x

 1  0  sin     1  1  sin   

 L / 2  2 2  L / 2  y    L/2  2 2  L / 2  y  

dq L

Ey 

 dE
k y

y





1 0

k y 

d  cos  

k y



1 0

cos  d 

Ey 

sin 

1 0

2k y

sin  1
because sin=-sin(-)

dE  y r
P 

y x

You can substitute for sin(1) if you want:
sin   1   L/2

dq L

L / 2

2

y

2

An insulating spherical shell has an inner radius a and outer radius b. The shell has a total charge Q and a uniform charge density. Find the magnitude of the electric field for r<a.
  q enclosed  E  dA    0  o

An insulating spherical shell has an inner radius a and outer radius b. The shell has a total charge Q and a uniform charge density. Find the magnitude of the electric field for a<r<b.
4 4 3 3 Q    b  a  3 3    q en clo sed  E  dA    o

E  4 r

2



4 3 4 3  r  a 3  3    o

An electron has a speed v. Calculate the magnitude and direction of an electric field that will stop this electron in a distance D.
E f  E i   E   W other i  f .

 V  Vf  Vi   Ed

U  qV

Do you understand that these E‘s have different meanings?

Use magnitudes if you have determined direction of E by other means.

Two equal positive charges Q are located at the base of an equilateral triangle with sides of length a. What is the potential at point P (see diagram)?

P

a

Q

Q

Three equal positive charges Q are located at the corners of an equilateral triangle with sides of length a. What is the potential energy of the charge located at point P (see diagram)?
P Q

a

Q

Q

Capacitors and Dielectrics
Capacitance
A capacitor is basically two parallel conducting plates with air or insulating material in between.

E V0 L V1

A capacitor doesn‘t have to look like metal plates.

Capacitor for use in high-performance audio systems.

The symbol representing a capacitor in an electric circuit looks like parallel plates. Here‘s the symbol for a battery, or an external potential.
+-

When a capacitor is connected to an external potential, charges flow onto the plates and create a potential difference between the plates. Capacitor plates build up charge.

V

-

The battery in this circuit has some voltage V. We haven‘t discussed what that means yet.

If the external potential is disconnected, charges remain on the plates, so capacitors are good for storing charge (and energy). Capacitors are also very good at releasing their stored charge all at once. The capacitors in your TV are so good at storing energy that touching the two terminals at the same time can be fatal, even though the TV may not have been used for months.

+-

+V

High-voltage TV capacitors are supposed to have ―bleeder resistors‖ that drain the charge away after the circuit is turned off. I wouldn‘t bet my life on it.
Graphic from http://www.feebleminds-gifs.com/.

assortment of capacitors

The charge acquired by each plate of a capacitor is Q=CV where C is the capacitance of the capacitor.
C Q V
C is always positive.

The unit of C is the farad but most capacitors have values of C ranging from picofarads to microfarads (pF to F).
micro 10-6, nano 10-9, pico 10-12

Parallel Plate Capacitance
We previously calculated the electric field between two parallel charged plates:
E   0  Q 0A .

-Q E V0 d

+Q

This is valid when the separation is small compared with the plate dimensions.

V1

A

We also showed that E and V are related:
V   
d 0

  d E  d s  E  dx  E d .
0

This lets us calculate C for a parallel plate capacitor.

C

Q V



Q Ed



Q  Q   d  0A 



0A d

Reminders:

C

Q V

Q is the magnitude of the charge on either plate. V is actually the magnitude of the potential difference between the plates.
C is always positive.

Clicker Time!

Parallel plate capacitance depends ―only‖ on geometry.
C 0A d

-Q

+Q

E
V0

This expression is approximate, and must be modified if the plates are small, or separated by a medium other than a vacuum.

d

V1

A

Coaxial Cylinder Capacitance
We can also calculate the capacitance of a cylindrical capacitor (made of coaxial cylinders).

 L

b

Δ V = Vb - Va = b


a

  E d s = -

b

E
a

r

dr

Gaussian surface
b

E =

2kλ r

ΔV = - 2k λ


a

dr

 b  = - 2 k λ ln   r  a 

r a Q

C=

Q ΔV

=

λL ΔV

=

λL b 2k λ ln   a 2 πε 0 L b ln   a

E ds -Q

C=

L b 2k ln   a

=

Lowercase c is capacitance per unit length: c =

2 πε 0 b n  a

Isolated Sphere Capacitance
An isolated sphere can be thought of as concentric spheres with the outer sphere at an infinite distance and zero potential.

We already know the potential outside a conducting sphere:
V  Q 4  0 r .

The potential at the surface of a charged sphere of radius R is
V  Q 4  0 R

so the capacitance at the surface of an isolated sphere is
C Q V  4  0 R .

Capacitance of Concentric Spheres
The hyperphysics web page shows how to calculate the capacitance of a spherical capacitor of charge Q.

In between the spheres
E  Q 4  0 r
2

V 

Q 4  0



b a

dr r
2

1 1   a  b 4  0   Q
I paid for a license to use hyperphysics.

C

Q V



4  0 1 1   a b  

This derivation looks ―just like‖ one of tomorrow‘s homework problems!

C

Q V



4  0 1 1   a b  

Let aR and b to get the capacitance of an isolated sphere.

Example: calculate the capacitance of a capacitor whose plates are 20 x 3 cm and are separated by a 1.0 mm air gap.
C 0A d

C

 8.85  10
 12

 12

  0.2  0.03 
d = 0.001

0.001

C  53  10 C  53 pF

F

area = 0.2 x 0.03

If you keep everything in SI (mks) units, the result is “automatically” in SI units.

Example: what is the charge on each plate if the capacitor is connected to a 12 volt* battery?
Q  CV
Q   53  10
 12

0V

 12 
C

V= 12V

Q  6.4  10

 10

+12 V

*Remember, it’s the potential difference that matters.

If you keep everything in SI (mks) units, the result is “automatically” in SI units.

Example: what is the electric field between the plates?
V d E 12 V 0.001 m

E 

0V

E d = 0.001

V= 12V

E  12000

V m

, " up."

+12 V

If you keep everything in SI (mks) units, the result is “automatically” in SI units.

Demo: Professor Tries to Avoid Spot-Welding His Fingers to the Terminals of a Capacitor While Demonstrating Energy Storage

Energy Storage in Capacitors
Let‘s calculate how much work it takes to charge a capacitor.

The work required for an external force to move a charge dq through a potential difference V is dW = dq V. From Q=CV:
dW  q C dq
This is the amount of charge on the capacitor during the time the charge dq is being moved.

+

V

-

dq +

We start with zero charge on the capacitor, and end up with Q, so
W 

+q

-q



Q 0

dW 



Q 0

q C

dq 

q

2

Q


0

Q

2

.

2C

2C

The work required to charge the capacitor is the amount of energy you get back when you discharge the capacitor (because the electric force is conservative). Thus, the work required to charge the capacitor is the potential energy stored in the capacitor.
U  Q
2

.

2C

Because C, Q, and V are related through Q=CV, there are three equivalent ways to write the potential energy.
U  Q
2



CV 2

2



QV 2

.

2C

U 

Q

2



CV 2

2



QV 2

.

2C

All three equations are valid; use the one most convenient for the problem at hand.

It is no accident that we use the symbol U for the energy stored in a capacitor. It is just another ―version‖ of electrical potential energy. You can use it in your energy conservation equations just like any other form of potential energy!

Example: a camera flash unit stores energy in a 150 F capacitor at 200 V. How much electric energy can be stored?
U  CV 2
2

U 

150  10

6

  200

2



2

U3J

If you keep everything in SI (mks) units, the result is “automatically” in SI units.

Example: compare the amount of energy stored in a capacitor with the amount of energy stored in a battery.

If a battery stores so much energy, what good are capacitors? Application: short pulse magnets at the National Magnet Laboratory 106 joules of energy are stored in capacitor banks, and released during a period of a few milliseconds. The enormous current produces incredibly high magnetic fields.

Capacitors and Dielectrics
Energy in Electric Fields
Energy is stored in the capacitor:
U  1 2
1  0A  2 U     Ed  2 d 
E d

CV

2

+

V

-

U 

1 2

2

+Q

-Q

area A

The ―volume of the capacitor‖ is Volume=Ad

Energy stored per unit volume (u):
1 u  2



1 2

0E

2

The energy is ―stored‖ in the electric field!

+

V E f

-

We‘ve gone from the concrete (electric charges experience forces)… …to the abstract (electric charges create electric fields)…
…to an abstraction within the abstraction (electric field contains energy).

+Q

-Q

area A

u 

1 2

0E

2

This is not a new ―kind‖ of energy. It‘s the electric potential energy resulting from the coulomb force between charged particles. Or you can think of it as the electric energy due to the field created by the charges. Same thing.

+

V E f

-

+Q

-Q

area A

Circuits Containing Capacitors in Parallel
Recall: this is the symbol representing a capacitor in an electric circuit.

And this is the symbol for a battery… …or this… …or this.

+-

Vab

Capacitors connected in parallel:
a

C1
C2 b

C3
+ V

The voltage drop from a to b must equal V. Vab = V = voltage drop across each individual capacitor.

Q=CV  Q1 = C1 V & Q2 = C2 V & Q3 = C3 V
a

C1
Q1

+
Q2

C2 -

C3
Q3 + -

V

Now imagine replacing the parallel combination of capacitors by a single equivalent capacitor.

a

Ceq
Q

By ―equivalent,‖ we mean ―stores the same total charge if the voltage is the same.‖ Q1 + Q2 + Q3 = Ceq V = Q

+ V

Summarizing the equations on the last slide: Q1 = C1 V Q2 = C2 V Q3 = C3 V
a

C1

C2
C3

b

Q1 + Q2 + Q3 = Ceq V

Using Q1 = C1V, etc., gives C1V + C2V + C3V = Ceq V
C1 + C2 + C3 = Ceq

+ V

(after dividing both sides by V)

Generalizing: Ceq = Ci (capacitors in parallel)

Capacitors connected in series:
C1 C2 C3

+ -

+Q V -Q

An amount of charge +Q flows from the battery to the left plate of C1. (Of course, the charge doesn‘t all flow at once).
An amount of charge -Q flows from the battery to the right plate of C3. Note that +Q and –Q must be the same in magnitude but of opposite sign.

The charges +Q and –Q attract equal and opposite charges to the other plates of their respective capacitors:
C1 +Q -Q

A

C2 +Q -Q + V

B

C3

+Q -Q

These equal and opposite charges came from the originally neutral circuit regions A and B.
Because region A must be neutral, there must be a charge +Q on the left plate of C2.

Because region B must be neutral, there must be a charge --Q on the right plate of C2.

Here‘s the circuit after the charges have moved and a steady state condition has been reached:
a
C1 +Q -Q V1

A

C2 +Q -Q V2 + V

B

C3 b

+Q -Q V3

The charges on C1, C2, and C3 are the same, and are Q = C1 V1 Q = C2 V2 Q = C3 V3
But we don‘t know V1, V2, and V3 yet.

We do know that Vab = V and also Vab = V1 + V2 + V3.

Let‘s replace the three capacitors by a single equivalent capacitor.
Ceq +Q -Q V + V

By ―equivalent‖ we mean V is the same as the total voltage drop across the three capacitors, and the amount of charge Q that flowed out of the battery is the same as when there were three capacitors. Q = Ceq V

Collecting equations: Q = C1 V1 Q = C2 V2 Q = C3 V3

Vab = V = V1 + V2 + V3.

Q = Ceq V Substituting for V1, V2, and V3:
Substituting for V:
V= Q C1 + Q C2 + Q C3

Q C eq 1 C eq

=

Q C1 1 C1

+

Q C2 1 C2

+

Q C3 1 C3

Dividing both sides by Q:

=

+

+

Generalizing:

OSE:

1 C eq

=


i

1 Ci

(capacitors in series)

Example: determine the capacitance of a single capacitor that will have the same effect as the combination shown. Use C1 = C2 = C3 = C.

C2

C1
C3

I don‘t see a series combination of capacitors, but I do see a parallel combination.
C23 = C2 + C3 = C + C = 2C

Now I see a series combination.
1 C eq 1 C eq = 1 C + = 1 2C 1 C1 = + 2 2C 1 C 23 + 1 2C = 3 2C

C23 = 2C

C1= C

C eq =

2 3

C

Demo: dielectrics (maybe!)

Dielectrics
If an insulating sheet (―dielectric‖) is placed between the plates of a capacitor, the capacitance increases by a factor , which depends on the material in the sheet.  is the dielectric constant of the material. In general, C = 0A / d.  is 1 for a vacuum, and  1 for air. (You can also define  = 0 and write C =  A / d).
C=   A d .

dielectric

The dielectric is the thin insulating sheet in between the plates of a capacitor.

dielectric

Any reasons to use a dielectric (other than to make your life more complicated)?
Lets you apply higher voltages (so more charge).

Lets you place the plates closer together (make d smaller). Increases the value of C because >1.
Q = CV   A C= d

Example: a parallel plate capacitor has an area of 10 cm2 and plate separation 5 mm. 300 V is applied between its plates. If neoprene is inserted between its plates, how much charge does the capacitor hold.
  A d

A=10 cm2

C=

C=

 6 .7   8 .8 5 × 1 0 -1 2  1 0 × 1 0 -4 
5 ×10
-11
-3

=6.7

C = 1.19  10 Q = CV

F

V=300 V d=5 mm
-11

Q = 1.19  10



 300   3.56  10 -9 

C = 3.56 nC

Example: how much charge would the capacitor on the previous page hold if the dielectric were air? A=10 cm2

The calculation is the same, except replace 6.7 by 1.

Or just divide the charge on the previous page by 6.7 to get.
Q = 0.53 nC

=1

V=300 V d=5 mm

Example
A capacitor connected as shown acquires a charge Q.

V=0

V

While the capacitor is still connected to the battery, a dielectric material is inserted. Will Q increase, decrease, or stay the same?
Why?

V

Example: find the energy stored in the capacitor on slide 21.
C = 1.19  10
1 2 U= 1
-11

F

A=10 cm2

U=

C  V 

2

 2

1.19  10

-11



 300 

2

=6.7

U = 5.36  10 J

-7

V=300 V d=5 mm

Example: the battery is now disconnected. What are the charge, capacitance, and energy stored in the capacitor? The voltage drop is unchanged, so the charge, capacitance, and energy stored are unchanged.
Q = 3.56 nC C = 1.19  10
-11

A=10 cm2

=6.7
F

U = 5.36  10 J

-7

V=300 V d=5 mm

Example: the dielectric is removed without changing the plate separation. What are the capacitance, charge, potential difference, and energy stored in the capacitor?

A=10 cm2
C= A d

C=

 8 .8 5 × 1 0  1 0 × 1 0 
-1 2 -4

=6.7 V=? V=300 V d=5 mm

5 ×10
-12

-3

C = 1.78  10

F

Example: the dielectric is removed without changing the plate separation. What are the capacitance, charge, potential difference, and energy stored in the capacitor?

The charge remains unchanged, because there is nowhere for it to go.
Q = 3.56 nC

A=10 cm2

V=? d=5 mm

Example: the dielectric is removed without changing the plate separation. What are the capacitance, charge, potential difference, and energy stored in the capacitor?

Knowing C and Q we can calculate the new voltage difference.
V = Q C =

A=10 cm2

 3 .5 6  1 0 1 .7 8  1 0

-9

-12

 
V=? d=5 mm

V = 2020 V

Example: the dielectric is removed without changing the plate separation. What are the capacitance, charge, potential difference, and energy stored in the capacitor?

A=10 cm2
U= 1 2 U= 1 C  V 
2

1.78  10   2020  2
-12

2

U = 3.63  10 J

-6

V=2020 V d=5 mm

U be fo re = 5 .3 6  1 0 J

-7

U a fte r = 3 .6 3  1 0 J

-6

U a fte r U b e fo re

= 6 .7

Huh?? The energy stored increases by a factor of ??
Sure. It took work to remove the dielectric. The stored energy increased by the amount of work done.
 U = W external

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