Physics 1502: Lecture 4 Today’s Agenda Announcements: –Lectures posted on: rcote/ rcote/ –HW assignments, solutions.

Physics 1502: Lecture 4Today’s Agenda

• Announcements:– Lectures posted on:

www.phys.uconn.edu/~rcote/

– HW assignments, solutions etc.

• Homework #1:Homework #1:– On Masterphysics today: due next FridayOn Masterphysics today: due next Friday

– Go to masteringphysics.com and register

– Course ID: MPCOTE33308

• Labs: Begin next week

Today’s Topic :• End of Chapter 21: Gauss’s Law

– Motivation & Definition– Coulomb's Law as a consequence of Gauss' Law– Charges on Insulators:

» Where are they?

• Chapter 22: Electric potential– Definition– How to compute it

Karl Friedrich Gauss(1777-1855)

€

ε0

r E • d

r S ∫ = ε0Φ = qenclosed

+ + + + ++ + + +x

y

+ + + + + + + + + + +

Infinite Line of Charge

• Symmetry E field must be to line and can only depend on distance from line

NOTE: we have obtained here the same result as we did last lecture using Coulomb’s Law. The symmetry makes today’s derivation easier!

Er

• Therefore, CHOOSE Gaussian surface to be a cylinder of radius r and length h aligned with the x-axis.

• Apply Gauss' Law:

AND q = h

h

Er

+ + + + + ++ + +

0 SdE• On the ends,

E dS rhE 2• On the barrel,

€

E =λ

2πε0r

Gauss’ Law: Help for the Problems• How to do practically all of the homework problems

• What Can You Do With This??

If you have (a) spherical, (b) cylindrical, or (c) planar symmetry AND:

• If you know the charge (RHS), you can calculate the electric field (LHS)

• If you know the field (LHS: usually because E=0 inside conductor), you can calculate the charge (RHS).

• Gauss’ Law is ALWAYS VALID!!

Application of Gauss’ Law:

RHS: q = ALL charge inside cylinder=A

• Planar Symmetry: Gaussian surface = Cylinder of area A

RHS: q = ALL charge inside radius r

• Spherical Symmetry: Gaussian surface = Sphere of radius r

RHS: q = ALL charge inside radius r, length L

• Cylindrical Symmetry: Gaussian surface = Cylinder of radius r

LHS:

LHS:

LHS:

Insulators vs. Conductors • Insulators – wood, rubber, styrofoam, most ceramics, etc.

• Conductors – copper, gold, exotic ceramics, etc.

• Sometimes just called metals

• Insulators – charges cannot move.

– Will usually be evenly spread throughout object

• Conductors – charges free to move.

– on isolated conductors all charges move to surface.

+++++++

-------

E+++++++

-------

E

Conductors vs. Insulators

++++++

------

+++++++

-------

E

+-

+

+-

-

- +

-+

+

-+ +

--

+

-+

+

-

-

- +

- +

- +

- +

- +

- +

- +

- +

- +

- + - +

- +

- +

- + - +- +

+++++++

-------

E

Ein = 0 Ein < E

Hollow conductors

Conductors & Insulators

• How do the charges move in a conductor??

• Hollow conducting sphere

Charge the inside, all of this charge moves to the outside.

• Consider how charge is carried on macroscopic objects. • We will make the simplifying assumption that there are only two kinds of objects in the world:

• Insulators.. In these materials, once they are charged, the charges ARE NOT FREE TO MOVE. Plastics, glass, and other “bad conductors of electricity” are good examples of insulators.

• Conductors.. In these materials, the charges ARE FREE TO MOVE. Metals are good examples of conductors.

Conductors vs. Insulators

+-

+

+-

-

-

+

-

-

- -

++

- +

+-- -

++++++

-

-- -++++++

- +

- + - +- +

++++++- +

- +

- +

- +

++++++- +

- +

- +

- +

- +

- + - +- +

Charges on a Conductor• Why do the charges always move to the surface of

a conductor ? – Gauss’ Law tells us!!

– E = 0 inside a conductor when in equilibrium (electrostatics) !

» Why?

If E 0, then charges would have forces on them and they would move !

• Therefore from Gauss' Law, the charge on a conductor must only reside on the surface(s) !

Infinite conductingplane

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

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

Conducting sphere

++

+

++

+

++

1

Lecture 4, ACT 1Consider the following two topologies:

1A • Compare the electric field at point X in cases A and B:

(a) 1 < (b) 1 = (c) 1 >

(a) EA < EB (b) EA = EB (c) EA > EB

1B • What is the surface charge density 1 on the inner surface of the conducting shell in case A?

E

2

A) A solid non-conducting sphere carries a total charge Q = -3 C distributed evenly throughout. It is surrounded by an uncharged conducting spherical shell.

B) Same as (A) but conducting shell removed.

1

-Q

Lecture 4, ACT 2• A line charge (C/m) is placed along the axis

of an uncharged conducting cylinder of inner radius ri = a, and outer radius ro = b as shown. – What is the value of the charge density o

(C/m2) on the outer surface of the cylinder?

(a) (b) (c)

a

b

Electric Potential

qA

C

B

rA

Br

path independence equipotentials

R

R

R r

VQ

4ε rQ

4ε R

Overview

• Introduce Concept of Electric Potential– Is it well-defined? i.e. is Electric Potential a property of the space

as is the Electric Field?

• Calculating Electric Potentials– Charged Spherical Shell

– N point charges

– Electric Dipole

• Can we determine the Electric Field if we know the Electric Potential?

Text Reference: Chapter 22

Electric Potential• Suppose charge q0 is moved from pt

A to pt B through a region of space described by electric field E.

• Since there will be a force on the charge due to E, a certain amount of work WAB will have to be done to accomplish this task. We define the electric potential difference as:

• Is this a good definition?

• Is VB - VA independent of q0?

• Is VB - VA independent of path?

A B

q0E

Independent of Charge?

• To move a charge in an E field, we must supply a force just equal and opposite to that experienced by the charge due to the E field.

A B

q0 EFelec

Fwe supply = -Felec

1

Lecture 4, ACT 3• A single charge ( Q = -1C) is fixed at the origin.

Define point A at x = + 5m and point B at x = +2m.

– What is the sign of the potential difference between A and B? (VAB VB - VA )

(a) VAB < (b) VAB = (c) VAB >

x-1C AB

Independent of Path?

A B

q0 EFelec

-Felec

• This equation also serves as the definition for the potential difference VB - VA.

•The integral is the sum of the tangential (to the path) component of the electric field along a path from A to B.

•The question now is: Does this integral depend upon the exact path chosen to move from A to B?

•If it does, we have a lousy definition. • Hopefully, it doesn’t. • It doesn’t. But, don’t take our word, see appendix and following example.

Does it really work?• Consider case of constant

field:– Direct: A - B

• Long way round: A - C - B

• So here we have at least one example of a case in which the integral is the same for BOTH paths.

A

CB

Eh rdl

Electric Potential• Define the electric potential of a point in space as the potential

difference between that point and a reference point.

• a good reference point is infinity ... we typically set V = 0

• the electric potential is then defined as:

• for a point charge, the formula is:

Potential from charged spherical shell

• E Fields (from Gauss' Law)

• Potentials

• r > R:

• r < R:

R

R

Q4ε R

R r

VQ

4ε r

E = 0• r < R:

E = 1

4ε Qr 2

• r > R:

Potential from N charges

The potential from a collection of N charges is just the algebraic sum of the potential due to each charge separately.

xr1

r2 r3

q1

q3

q2

Electric Dipole z

a

a

+q

-q

r

r1r2

The potential is much easier to calculate than the field since it is an algebraic sum of 2 scalar terms.

• Rewrite this for special case r>>a:

Can we use this potential somehow to calculate the E field of a dipole?

(remember how messy the direct calculation was?)

r2-r1

Appendix: Independent of Path?

qA

C

B

rA

Br

E

• We want to evaluate potential difference from A to B

qA

B

rA

Br

E

• What path should we choose to evaluate the integral?.

• If we choose straight line, the integral is difficult to evaluate.

• Magnitude different at each pt along line.• Angle between E and path is different at each pt along line.

• If we choose path ACB as shown, our calculation is much easier!

• From A to C, E is perpendicular to the path. ie

• From A to C, E is perpendicular to the path. ie

Appendix: Independent of Path?• Evaluate potential difference from

A to B along path ACB.

qA

C

B

rA

Br

E

Evaluate the integral:

by definition:

Appendix: Independent of Path?

• How general is this result?• Consider the approximation to the straight

path from A->B (white arrow) = 2 arcs (radii = r1 and r2) plus the 3 connecting radial pieces.

qA

B

This is the same result as above!!The straight line path is better

approximated by Increasing the number of arcs and radial pieces.

qA

B

rA

Br

C

r1

r2

• For the 2 arcs + 3 radials path:

Appendix: Independent of Path?

• Consider any path from A to B as being made up of a succession of arc plus radial parts as above. The work along the arcs will always be 0, leaving just the sum of the radial parts. All inner sums will cancel, leaving just the initial and final radii as above.. Therefore it's general!

qA

B

r1

r2