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W04D1 Electric Potential and Gauss’ Law

Equipotential Lines

Today’s Reading Assignment Course Notes: Sections 3.3-3.4, 4.4-4.6. 4.10.5

Announcements

Exam One Thursday Feb 28 7:30-9:30 pm Room Assignments (See Stellar webpage announcements) Review Tuesday Feb 26 from 9-11 pm in 26-152 PS 3 due Tuesday Tues Feb 26 at 9 pm in boxes outside 32-082 or 26-152

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Outline

Review E and V Deriving E from V

Using Gauss’s Law to find V from E

Equipotential Surfaces

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Electric Potential and Electric Field

Set of discrete charges: Continuous charges: If you already know electric field (Gauss’ Law) compute electric potential difference using

V (!r) !V (") = ke

d #q!r - !#rsource

$

VB !VA = !

!E

A

B

" # d !s

V (!r)!V (") = ke

qi!r - !rii#

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E Field and Potential: Effects

!F = q

!E

If you put a charged particle, (charge q), in a field:

Wext =!U = q!V

To move a charged particle, (charge q), in a field and the particle does not change its kinetic energy then:

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Concept Question: Two Point Charges

The work done in moving a positively charged object that starts from rest at infinity and ends at rest at the point P midway between two charges of magnitude +Q and –Q 1.  is positive. 2.  is negative. 3.  is zero. 4.  can not be determined – not enough info is given.

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E Field and Potential: Creating

A point charge q creates a field and potential around it: Use superposition for systems of charges

V (!r) !V (") = ke

d #q!r - !#rsource

$

!E = ke

qr 2 r

V (r) !V (") = ke

qr

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Using Gauss’s Law to find Electric Potential from Electric Field

If the charge distribution has a lot of symmetry, we use Gauss’s Law to calculate the electric field and then calculate the electric potential V using

B

B AA

V V d− = − ⋅∫E s

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Group Problem: Coaxial Cylinders

A very long thin uniformly charged cylindrical shell (length h and radius a) carrying a positive charge +Q is surrounded by a thin uniformly charged cylindrical shell (length h and radius a ) with negative charge -Q , as shown in the figure. You may ignore edge effects. Find V(b) – V(a).

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Worked Example: Spherical Shells These two spherical shells have equal but opposite charge. Find for the regions (i)  b < r (ii)  a < r < b (iii)  0 < r < a Choose V (!) = 0

V (r) !V (")

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Electric Potential for Nested Shells

!E =

Q4!"0r

2 r, a < r < b

0, elsewhere

#

$%

&%%

From Gauss’s Law

V (r) !V (")

= 0! = ! 0

"

r

# dr = 0

VB !VA = !

!E

A

B

" # d !sUse

Region 1: r > b

r

No field No change in V!

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Electric Potential for Nested Shells Region 2: a < r < b

V (r) !V (r = b)= 0

!"# $# = ! dr Q4"#0r

2b

r

$

= Q

4!"0r r=b

r

= 1

4!"0

Q 1r# 1

b$%&

'()

r

Electric field is just a point charge. Electric potential is DIFFERENT – surroundings matter

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Electric Potential for Nested Shells Region 3: r < a

V (r) ! V (a)

= kQ 1a!

1b

"#$

%&'

! = ! dr 0a

r

( = 0

V (r) =V (a) = Q

4!"0

1a# 1

b$%&

'()

r

Again, potential is CONSTANT since E = 0, but the potential is NOT ZERO for r < a.

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Group Problem: Charge Slab

Infinite slab of thickness 2d, centered at x = 0 with uniform charge density . Find !

V (xA ) !V (0) ; xA > d

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Deriving E from V

!!s = !x i

A = (x,y,z), B=(x+Δx,y,z)

Ex ! " #V

#x$ " %V

%xEx = Rate of change in V with

y and z held constant

!V = "

!E

A

B

# $d !s

!V = "

!E

( x ,y ,z )

( x+!x ,y ,z )

# $d !s % "!E $ !!s = "

!E $(!x i) = "Ex!x

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Gradient (del) operator:

!! " #

#xi + #

#yj + #

#zk

If we do all coordinates:

!E = !

!"V

Deriving E from V

!E = ! "V

"xi + "V

"yj + "V

"zk

#$%

&'(

= ! "

"xi + "

"yj + "

"zk

#$%

&'(

V

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Concept Question: E from V Consider the point-like charged objects arranged in the figure below. The electric potential difference between the point P and infinity and is

V (P) = ! kQ aFrom that can you derive E(P)?

1.  Yes, its kQ/a2 (up) 2.  Yes, its kQ/a2 (down) 3.  Yes in theory, but I don’t know how to take a

gradient 4.  No, you can’t get E(P) from V(P)

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Group Problem: E from V Consider two point like charged objects with charge –Q located at the origin and +Q located at the point (0,a). (a) Find the electric potential V(x,y)

at the point P located at (x,y).

(b) Find the x-and y-components of the electric field at the point P using

Ex (x, y) = ! "V

"x, Ey (x, y) = ! "V

"y

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Concept Question: E from V

The graph above shows a potential V as a function of x. The magnitude of the electric field for x > 0 is

1.  larger than that for x < 0 2.  smaller than that for x < 0 3.  equal to that for x < 0

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Concept Question: E from V

The above shows potential V(x). Which is true?

1.  Ex > 0 is positive and Ex < 0 is positive 2.  Ex > 0 is positive and Ex < 0 is negative 3.  Ex > 0 is negative and Ex < 0 is negative 4.  Ex > 0 is negative and Ex < 0 is positive

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Group Problem: E from V

A potential V(x,y,z) is plotted above. It does not depend on x or y. What is the electric field everywhere? Are there charges anywhere? What sign?

-5 0 50

5

10Po

tent

ial (

V)

Z Position (mm)

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Equipotentials

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Topographic Maps

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Equipotential Curves: Two Dimensions

All points on equipotential curve are at same potential. Each curve represented by V(x,y) = constant

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Direction of Electric Field E

E is perpendicular to all equipotentials

Constant E field Point Charge Electric dipole

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Direction of Electric Field E

http://web.mit.edu/viz/EM/visualizations/electrostatics/InteractingCharges/zoo/zoo.htm

E is perpendicular to all equipotentials Field of 4 charges Equipotentials of 4 charges

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Properties of Equipotentials

E field lines point from high to low potential E field lines perpendicular to equipotentials

E field has no component along equipotential The electrostatic force does zero work to

move a charged particle along equipotential

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Group Problem: Equipotential Game Open applet, play with, answer

questions on worksheet and hand in answers

http://web.mit.edu/viz/EM/visualizations/electrostatics/EandV/VfromE/VfromE.htm