Welcome…

…to Electrostatics

Outline

1. Coulomb’s Law2. The Electric Field

- Examples3. Gauss Law

- Examples4. Conductors in Electric Field

Coulomb’s Law

1 22

12

q qF k12 r

Coulomb’s law gives the force (in Newtons) between charges q1

and q2, where r12 is the distance in meters between the charges,

and k=9x109 N·m2/C2.

Coulomb’s law quantifies the magnitude of the electrostatic force.

Force is a vector quantity. The equation on the previous slide gives the magnitude of the force.

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 1/40,

where 0=8.85x10-12 C2/N·m2.

Coulomb’s Law

212

21

012 4

1

r

qqF

Coulomb’s Law

• If q1 and q2 are located at points having position vectors r1 and r2 then the Force F12 is given by;

1212

212

21

012 124

1

rrR

aF

Rr

qq

The equation is valid for point charges. If the charged objects are spherical and the charge is uniformly distributed, r12 is the distance between the centers of the spheres.

If more than one charge is involved, the net force is the vector sum of all forces (superposition). For objects with complex shapes, you must add up all the forces acting on each separate charge (turns into calculus!).

+ -

r12

++

+-

--

Solving Problems Involving Coulomb’s Law and Vectors

x

y

Q2=+50C

Q3=+65C

Q1=-86C

52 cm

60 cm

30

cm

=30º

Example: Calculate the net electrostatic force on charge Q3 due to the charges Q1 and Q2.

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.

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.

Draw and label forces (only those on Q3).

Draw components of forces which are not along axes.

x

y

Q2=+50C

Q3=+65C

Q1=-86C

52 cm

60 cm

30

cm

=30º

F31

F32Draw a representative sketch.

Draw and label relevant quantities.

Draw axes, showing origin and directions.

Step 1: Diagram

1 22

12

q qF k12 r

“Do I have to put in the absolute value signs?”

x

y

Q2=+50C

Q3=+65C

Q1=-86C

52 cm

60 cm

30

cm

=30º

F31

F32

Step 2: Starting Equation

3 2232

Q QF k ,32 r

repulsive

3 2232

Q QF k32, y r

F 032, x

(from diagram)

F32,y = 330 N and F32,x = 0 N.

x

y

Q2=+50C

Q3=+65C

Q1=-86C

52 cm

r31 =60 cm

r 32=

30

cm

=30º

F31

F32

Step 3: Replace Generic Quantities by Specifics

3 1231

Q QF k ,31 r

attractive

3 1231

Q QF k cos31, x r

You would get F31,x = +120 N and F31,y = -70 N.

(- sign comes from diagram)3 1231

Q QF k sin31, y r

(+ sign comes from diagram)

x

y

Q2=+50C

Q3=+65C

Q1=-86C

=30º

F31

F32

Step 3 (continued)

r 32=

30

cm r

31 =60 cm

52 cm

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.

F3The net force is the vector sum of all the forces on Q3.

x

y

Q2=+50C

Q3=+65C

Q1=-86C

52 cm

60 cm

30

cm

=30º

F31

F32

Step 3: Complete the Math

Many Point Charges

• If we have N point charges Q1, Q2… QN, located at r1, r1… rN the vector sum of the forces exerted on Q by each of the charges is given by

N

k k

kkQQ

13

0

)(

4 rr

rrF

Coulomb's Law quantifies the interaction between charged particles.

+ -

Coulomb’s Law:The Big Picture

Q1 Q2

r121 2

20 12

q q1F = ,12 4πε r

Coulomb’s Law was discovered through decades of experiment.

By itself, it is just “useful." Is it part of something bigger?

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?”

Faraday, beginning in the 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.

+

+-

like charges

repel

unlike charges attract

F12

F21

F31

F13

The Electric Field

The Electric Field

We define the electric field by the force it exerts on a test charge q0:

0

0

FE =

q

By convention the direction of the electric field is the direction of the force exerted on a POSITIVE test charge. The absence of absolute value signs around q0 means you must include the sign of q0 in your work.

If the test charge is "too big" it perturbs the electric field, so the “correct” definition is

0

0

q 00

FE = lim

q

Any time you know the electric field, you can use this equation to calculate the force on a charged particle in that electric field.

You won’t be required to use this version of the equation.

F = qE

The Electric Field

The units of electric field are Newtons/Coulomb.

0

0

F NE = =

q C

Later you will learn that the units of electric field can also be expressed as volts/meter:

N VE = =

C m

The electric field exists independent of whether there is a charged particle around to “feel” it.

+Remember: 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).

The Electric FieldDue to a Point Charge

Coulomb's law says

... which tells us the electric field due to a point charge q is

1 22

12

q qF =k ,12 r

This is your third starting equation.

q 2

qE =k , away from +

r

…or just…

2

qE=k

r

The equation for the electric field of a point charge then becomes:

2

qˆE=k r

r

We define as a unit vector from the source point to the field point:

r̂

+source point

field point

r̂

You may start with either equation for the electric field (this one or the one on the previous slide). But don’t use this one unless you REALLY know what you are doing!

Many Point Charges

• If we have N point charges Q1, Q2… QN, located at r1, r1… rN the vector sum of the Electric Field E by each of the charges is given by

N

k k

kkQ

13

0

)(

4

1

rr

rrE

Motion of a Charged Particlein a Uniform Electric Field

A charged particle in an electric field experiences a force, and if it is free to move, an acceleration.

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

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

-F

If the only force is due to the electric field, then

F ma qE.

If E is constant, then a is constant, and you can use the equations of kinematics.

E

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.

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

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

E

x

y

-

v0

The Electric FieldDue to a Collection of Point Charges

The electric field due to a small "chunk" q of charge is

20

1 qE = r

4πε r

unit vector from q to wherever you want to calculate E

The electric field due to collection of "chunks" of charge is

iii 2

i i0 i

1 qE = E = r

4πε r

As qdq0, the sum becomes an integral.

Charge Distributions

• Continious charge distributions are given by;

• Do not confuse ρ with radial distance

If charge is distributed along a straight line segment parallel to the z-axis, the amount of charge dq on a segment of length dz is ρ dz.

The Charge elemen is given by;

The total charge is given by;

Line Charge

The electric field at point P due to the charge dq is

I’m assuming positively charged objects in these “distribution of charges” slides.

Line Charge

2 20 0

1 dq 1 dxdE = r' = r'

4πε r' 4πε r'

The field point is denoted by (x,y,z) while the source point is denoted by (x’,y’,z’)

Line Charge

Definitions

Line Charge

• Using given definitions we have;

• For a finite line charge;

Line Charge

• For an infinite line charge α1 and α2 are equal to π/2 and - π/2 respectively, the z component is canceled and the equation becomes

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

x

y

area = dS

ρs

charge dQ = ρs dS

Surface Charge

The electric field at P due to the charge dq is

ydE

x

P

r’

ra

Surface Charge

The net electric field at P due to the entire surface of charge is

dE

x

P

r’

ra

Surface Charge

R

s

S

r

dSaE

20 '4

1

Surface Charge

• Definitions

• And the E field becomes;

Surface Charge

• After evaluating the previous integration;

• Or a more general formula is

• Note that the Electric Field is independent of the distance between the sheet and the point of observation P

an is a unit vector normal to the sheet

x

zP

r’r'

20 V

1 (x, y, z) dVE = r' .

4πε r'

The net electric field at P due to a three-dimensional distribution of charge is…

y

E

Volume Charge

Volume Charge

• For a sphere with radius a centered at the origin

• The E field due to elementary volume charge is given by;

Volume Charge

• Due to the symmetry of the charge distribution, the contributions to Ex or Ey add up to zero

Volume Charge

• Definitions

• And the resulting expression for E field is;

Summarizing:

20

1 λ dxE = r' .

4πε r'

20 S

1 dSE = r' .

4πε r'

20 V

1 dVE = r' .

4πε r'

Charge distributed along a line:

Charge distributed over a surface:

Charge distributed inside a volume:

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

The Electric FieldDue to a Continuous Charge

Distribution(worked examples)

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.

P x

y

d L

Let’s put the origin at P. The linear charge density and Q are related by

Q = and Q = L

L

Let’s assume Q is positive.

P x

y

d L

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 x dx dQ = dx

2 2

dq dxdE = k k

x x

Note: dE is in the –x direction. dE is the magnitude of dE. I’ve used the fact that Q>0 (so dq=0) to eliminate the absolute value signs in the starting equation.

P x

y

d L

dE x dx dQ = dx

d Ld+L d+L d+L

x 2 2d d dd

dx dx 1ˆ ˆ ˆE = dE = -k i = -k i = -k ix x x

1 1 d d L L kQˆ ˆ ˆ ˆE = -k i = -k i= -k i= - i

d L d d d L d d L d d L

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.

P

a

dQ

r

x0

dE

x

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

P

a

dQ

r

dE

x

2

dQdE=k

r

x 2

dQdE =k cos

r

2 20r = x a 0x

cosr

x0

0 0 0 0

x x 3/ 22 3 3 2 2ring ring ring 0

x x x kx QdQE dE k k dQ k Q

r r r r x a

Or, in general, on the ring axis x,ring 3/ 22 2

kxQE .

x a

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

No absolute value signs because Q is positive.

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.

P

r

dQ

xx0R

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

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.

0

ring 3/ 22 20

kx 2 rdrdE .

x r

Caution! I’ve switched the “meaning” of r!

P

r

dQ

xx0R

R

0x x 03/ 2 3/ 202 2 2 2

disc disc 0 0

kx 2 rdr 2r drE dE kx

x r x r

R1/ 22 20 0 0

x 0 1/ 22 20 0

0

x r x xE kx 2k

1/ 2 x x 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 to get0

1k

4

sheet0

E .2

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

I’ve been Really Nice and put this on your starting equation sheet. You don’t have to derive it for your homework!

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. Easier to use this link!

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, except that…

If the surface is tilted, fewer lines cut the surface.

EA

E

Electric Flux

E

A

The “amount of surface” perpendicular to the electric field is A cos .

A = A cos so E = EA = EA cos .

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

Because A is perpendicular to the surface, the amount of A parallel to the electric field is A cos .

Remember the dot product? AE E

Electric Flux

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…

dAE

A

Electric Flux

AE dE

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

E

…we count* lines going out as positive and lines going in as negative…

dAa surface integral, therefore a double integral

AE dE

Electric Flux

E E dA

E E dA

E E A

E EA cos

E EA

Use the simplest (easiest!) one that works.

Flat surface, E A, E constant over surface. Easy!

Flat surface, E not A, E constant over surface.

Flat surface, E not A, E constant over surface.

Surface not flat, E not uniform. Avoid, if possible.

Closed surface. Most general. Most complex.

If the surface is closed, you may be able to “break it up” into simple segments and still use E=E·A for each segment.

Electric Flux

Gauss’ Law

Gauss’ Law

Gauss's law states that the total electric flux through any closed surface is equal to the total charge enclosed by that surface.

E

Gauss’ Law

Differential form of Gauss’ Law

the volume charge density is the same as the divergence of the electric flux density.

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…

Example: use Gauss’ Law to calculate the electric field from an isolated point positive 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.

Point Charge Example

Point Charge Example

Starting with Gauss’s law, calculate the electric field due to an isolated point charge q.

qE

r dA

We choose a Gaussian surface that is a sphere of radius r centered on the point charge. Since the charge is positive the field is radial outward and everywhere perpendicular to the Gaussian surface.

Gauss’s law then gives:

Symmetry tells us that the field is constant on the Gaussian surface.

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.

Evaluate the surface integral (electric flux).

Determine the charge inside the Gaussian surface.

Solve for E.

Use symmetry to determine the direction of E on the Gaussian surface.

Example: use Gauss’ Law to calculate the electric field due to a long line of charge, with linear charge density .

Example: use Gauss’ Law to calculate the electric field due to an infinite sheet of charge, with surface charge density .

These are easy using Gauss’ Law (remember what a pain they were in the previous chapter). Study these examples and others in your text!

sheet0

E .2

line0

E .2 r

Worked Example 1Compute the electric flux through a cylinder with an axis parallel to the electricfield direction.

E

The flux through the curved surface is zero since E is perpendicular to dA there. For the ends, the surfaces are perpendicular to E, and E and A are parallel. Thus the flux through the left end (into the cylinder) is –EA, while the flux through right end (out of the cylinder) is +EA. Hence the net flux through the cylinder is zero.

A

Gauss’s LawGauss’s Law relates the electric flux through a closed surface withthe charge Qin inside that surface.

0

inE

QE dA

This is a useful tool for simply determining the electric field, but only for certain situations where the charge distribution is either rather simple or possesses a high degree of symmetry.

Worked Example 3An insulating sphere of radius a has a uniform charge density ρ and a total positive charge Q. Calculate the electric field outside the sphere.

a

Since the charge distribution is spherically symmetric we select a spherical Gaussian surface of radius r > a centered on the charged sphere. Since the charged sphere has a positive charge, the field will be directed radially outward. On the Gaussian sphere E is always parallel to dA, and is constant.Q

rE

dA

2Left side: 4E dA E dA E dA E r

0 0

Right side: inQ Q

22 2

0 0

14 or

4 e

Q Q QE r E k

r r

Worked Example 3 cont’d

a

Q

Find the electric field at a point inside the sphere.

Now we select a spherical Gaussian surface with radius r < a. Again the symmetry of the charge distribution allows us to simply evaluate the left side of Gauss’s law just as before.

r

The charge inside the Gaussian sphere is no longer Q. If we call the Gaussian sphere volume V’ then

2Left side: 4E dA E dA E dA E r

3

2

0 0

44

3inQ r

E r

34Right side:

3inQ V r

3

3 3230 00

4 1 but so

43 43 43

e

r Q Q QE r E r k r

a ar a

Worked Example 3 cont’d

2

3

We found for ,

and for ,

e

e

Qr a E k

rk Q

r a E ra

a

Q

Let’s plot this:E

ra

Conductors in Electrostatic Equilibrium

The electric field is zero everywhere inside the conductorAny net charge resides on the conductor’s surfaceThe electric field just outside a charged conductor is perpendicular to the conductor’s surface

By electrostatic equilibrium we mean a situation where there is no net motion of charge within the conductor

Conductors in Electrostatic Equilibrium

Why is this so?

If there was a field in the conductor the chargeswould accelerate under the action of the field.

The electric field is zero everywhere inside the conductor

++

++

++

++

++

++

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

Ein

E E

The charges in the conductor move creating an internal electric field that cancels the applied field on the inside of the conductor

The charge on the right is twice the magnitude of the charge on the left (and opposite in sign), so there are twice as many field lines, and they point towards the charge rather than away from it.

Combinations of charges. Note that, while the lines are less dense where the field is weaker, the field is not necessarily zero where there are no lines. In fact, there is only one point within the figures below where the field is zero – can you find it?

The electric field is always perpendicular to the surface of a conductor – if it weren’t, the charges would move along the surface.

The electric field is stronger where the surface is more sharply curved.

Summary

Two methods for calculating electric field Coulomb’s LawGauss’s Law

Gauss’s Law: Easy, elegant method for symmetric charge distributionsCoulomb’s Law: Other casesGauss’s Law and Coulomb’s Law are equivalent for electric fields produced by static charges

Worked Example 4Any net charge on an isolated conductor must reside on its surface and the electric field just outside a charged conductor is perpendicular to its surface (and has magnitude σ/ε0). Use Gauss’s law to show this.For an arbitrarily shaped conductor

we can draw a Gaussian surface inside the conductor. Since we have shown that the electric field inside an isolated conductor is zero, the field at every point on the Gaussian surface must be zero.

From Gauss’s law we then conclude that the net charge inside the Gaussian surface is zero. Since the surface can be made arbitrarily close to the surface of the conductor, any net charge must reside on the conductor’s surface.

0

inQE dA

Worked Example 4 cont’dWe can also use Gauss’s law to determine the electric field just outside the surface of a charged conductor. Assume the surface charge density is σ. Since the field inside the

conductor is zero there is no flux through the face of the cylinder inside the conductor. If E had a component along the surface of the conductor then the free charges would move under the action of the field creating surface currents. Thus E is perpendicular to the conductor’s surface, and the flux through the cylindrical surface must be zero. Consequently the net flux through the cylinder is EA and Gauss’s law gives:

0 0 0

orinE

Q AEA E

Worked Example 5A 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 find the electric field everywhere, and to determine the charge distribution on the spherical shell.

a

b

-Q First find the field for 0 < r < a

This is the same as Ex. 2 and is the field due to apoint charge with charge +2Q.

2

2e

QE k

r

Now find the field for a < r < bThe field must be zero inside a conductor in equilibrium. Thus from Gauss’s law Qin is zero. There is a + 2Q from the point charge so we must have Qa = -2Q on the inner surface of the spherical shell. Since the net charge on the shell is -Q we can get the charge on the outer surface from Qnet = Qa + Qb.

Qb= Qnet - Qa = -Q - (-2Q) = + Q.

+2Q

Worked Example 5 cont’d

a

b

-Q

+2Q

Find the field for r > bFrom the symmetry of the problem, the field in this region is radial and everywhere perpendicular to the spherical Gaussian surface. Furthermore, the field has the same value at every point on the Gaussian surface so the solution then proceeds exactly as in Ex. 2, but Qin=2Q-Q.

24E dA E dA E dA E r

Gauss’s law now gives:

22 2

0 0 0 0

2 14 or

4in

e

Q Q Q Q Q QE r E k

r r