TUTORIAL 7Discussion of Quiz 2

Solution of Electrostatics part 1

Quiz 2 - Question 1

! Postulations of ElectrostaticsPostulations of Electrostatics

3

Brief Review

1. What is field ? Scalar field and vector field

2. Definitions and physical meanings of gradient,

divergence and curl of a ______ field

3. Divergence Theorem:

s cd d!" #$ $! !"( A) s A l

4. Stoke’s Theorem:

v sdv d! #$ $! !"A A s

5. Two Null Identities:

! " #$$$$$$$ ! " #V!" ! # ! !" #! A

4

Postulates of Electrostatics in Free Space

• is the permittivity of free space

• Electric fields mentioned here only due to static charges in free space

• Physical meaning of them:

(1) Implies that a static electric field is not solenoidal

(2) Asserts that static electric fields are irrotational.

is a form of Gauss’Law

means the scalar line integral of the static electric field

intensity around any close path vanishes. Kichhoff Voltage’s Law

#

%&''()(*+&,-$'.)/

$$$$$$$$$$

#

%&

! #

!" #

!

:

(1) E

(2) E

#&

#c

d #$ !" E l

#

0*+(1),-$'.)/2

$$$$$$$$$$

#

s

c

Qd

d

&#

#

$

$

E s

E l

!

!

"

"

#

$s

Qd

&#$ !" E s

! Static Electric field is not solenoidal

! Static Electric field is irrotational

! Electric potential - the line integral of E

Postulations of Electrostatics

3

Brief Review

1. What is field ? Scalar field and vector field

2. Definitions and physical meanings of gradient,

divergence and curl of a ______ field

3. Divergence Theorem:

s cd d!" #$ $! !"( A) s A l

4. Stoke’s Theorem:

v sdv d! #$ $! !"A A s

5. Two Null Identities:

! " #$$$$$$$ ! " #V!" ! # ! !" #! A

4

Postulates of Electrostatics in Free Space

• is the permittivity of free space

• Electric fields mentioned here only due to static charges in free space

• Physical meaning of them:

(1) Implies that a static electric field is not solenoidal

(2) Asserts that static electric fields are irrotational.

is a form of Gauss’Law

means the scalar line integral of the static electric field

intensity around any close path vanishes. Kichhoff Voltage’s Law

#

%&''()(*+&,-$'.)/

$$$$$$$$$$

#

%&

! #

!" #

!

:

(1) E

(2) E

#&

#c

d #$ !" E l

#

0*+(1),-$'.)/2

$$$$$$$$$$

#

s

c

Qd

d

&#

#

$

$

E s

E l

!

!

"

"

#

$s

Qd

&#$ !" E s

! The Scalar line integral of the static electric field intensity around any close path vanishes.

z

yx

I1

I2

I3

r’

Quiz 2 - Question 1

Summary

G. Sciolla – MIT 8.022 – Lecture 4

14

Curl in cartesian coordinates (2)

! Combining this result with definition of curl:

! "

0

0 0

ˆ

)

curl C

A square yz x

x y yz

F ds F n F ds FFA

F x y y zFF

F ds y z

!

" !

" !

#

$ % &' (&$ ) * * +, - ." " & &&' (& / 0$ * +- .$ & &/ 01

2 2

2

! !"#! ! !

" "!

! !"

$ $

$

ˆ ˆ ˆ

ˆ ˆ ˆcurl y yx xz z

x y z

x y z

F FF FF FF x y z F

y z z x x y x y z

F F F

& &' ( ' (& && & & & &' ( * + 3 + 3 + % 4- . - .- .& & & & & & & & &/ 0/ 0 / 0

! !

l: easy to calculate! 13

Similar results orienting the rectangles in // (xz) and (xy) planes

square

lim

(curl lim

y z " "

% 5

This is the usable expression for the cur

Summary of vector calculus in

! Gradient:

!

! Divergence:

!

! l i

! Curl:

!

!

, , x y z

6 6' (& & &

5 % - .& & &/ 0

E 6 !

yx z FF F

F x y z

&& &5 * 3 3

& & &

! "

S V E dA * 52 2 !! ! " "

4E 785 *

! "

Purcell Chapter 2

A =

C F ds F dA2 2

!! !!" "$

F F* 54

! !

0E54 *

!

G. Sciolla – MIT 8.022 – Lecture 4

electrostatics (1)

In E&M:

Gauss’s theorem:

In E&M: Gauss’ aw in different al form

Stoke’s theorem:

In E&M:

* +5

EdV

curl

curl

7

11

G. Sciolla – MIT 8.022 – Lecture 6 21

Thoughts about ! and E

! The potential !"is always continuous

! E is not always continuous: it can “jump”

! When we have surface charge distributions

! Remember problem #1 in Pset 2

" When solving problems always check for consistency!

G. Sciolla – MIT 8.022 – Lecture 6 22

Summary

#(x,y,z)

$(x,y,z) !(x,y,z)

Gauss

’s law

(int)

V

d q

r! % &

2

2 11

E ds! !' % '&!""

#

2 4 ! (#) % '

E !% ')"

4 E (#) %"

#

Quiz 2 - Question 2

1. Total charge

Quiz 2 - Question 2

!

Q(r) = "(r r )

V

### dV

= rd(4

3$r

3)

0

r

#

= 4$r3dr

0

r

#= $r

4

!

Q(r) = "(r r )

V

### dV

= rd(4

3$r

3)

0

1

#

= 4$r3dr

0

1

#= $

2. Electric field - direction

Quiz 2 - Question 2

!

v E (

v r ) ="V (

v r )

=#V (

v r )

#xˆ x +

#V (v r )

#yˆ y +

#V (v r )

#zˆ z

=dV

dr(#r

#xˆ x ) +

dV

dr(#r

#yˆ y ) +

dV

dr(#r

#zˆ z )

=dV

dr

xˆ x + yˆ y + zˆ z

r=

dV

dr

ˆ a

r

2. Electric field - magnitude

3. Electric potential

Quiz 2 - Question 2

!

r E " d

r s

S## = E

0(r)ds

S##

= 4$r2E0(r)

=Q

%0

!

V (0) "V (r) =r E # d

r r

0

r

$

Outline forSolution of electrostatics

! Poisson’s and Laplace’s equation! Uniqueness theorem! Method of images! Method of separation of variables (Next tutorial)

What are we dealing with?

Given:

Charge source

Background material

Boundary condition

Question:

Find the E and V

Governing equation:

!

"2V = #

$

%

Q

!

V

Poisson’s equation! Laplacian

! Laplacian stands for the divergence of the gradient of

! Poisson Equation

! is a second order differential equation holds at every points in space.

! Laplace Equation

! is the governing equation for problems involving a set of conductors

!

"2V =" #"V

!

"2V = (

r a x

#

#x+

r a y

#

#y+

r a y

#

#y) $ (

r a x#V

#x+

r a y#V

#y+

r a y#V

#y)

=# 2V

#x2

+# 2V

#y2

+# 2V

#y2

!

"2V = #

$

%

!

"2V = 0

Uniqueness of Electrostatic solutions

! A solution of Poisson’s equation that satisfies the given boundary conditions is an unique solution.

! A solution of an electrostatic problem satisfying its boundary conditions is the only possible solution

! Method of image : Field E is uniquely defined by its normal components over the surface which confines this region.

Method of images

! Replace the original boundary by appropriate image charges in lieu of formal solution of Poisson’s or Laplace’s equation.

! Reminder:

! Image charges are virtual charges

! Use image charges to set up the boundary conditions.

! The images charge should be located outside the region in which the field is to be determined.

Boundary condition for one typical case

! Laplace’s equation in Cartesian coordinate system

! Boundary conditions

Example 1

! Determine the system of image charges that will replace the conduction boundaries that are maintained at zero potential for

! a) A point charge Q located between two large, grounded, parallel conducting planes as shown in fig (a)

d

d/3

V=0

V=0

In the first iteration

We will have

Q1’=-Q at 2/3d up of plane S1

Q2’=-Q at 1/3 below the plane S2

In the second iteration

We will have

Q1’’=+Q at d+1/3d up of plane S1

Q2’’=+Q at d+2/3d up of plane S2

Q1

Q2

Q1’

Q2’

Q1’’

Q2’’

Finally, at below of plane S2

and at up of plane S1

!

("1)n+1Q

!

("1)n+1Q

!

[2n +1

3("1)

n]d

!

[2n +2

3("1)

n]d

Example 1! Determine the system of image charges that will replace the conduction boundaries that are maintained at zero potential for

! b) An infinite line charge located midway between two large, intersecting conducting planes forming a 60 degree angle

V=0

V=0

60

When a point charge located between two paralleled conducting planes, the number of image charges goes to infinite.

When a point charge located between two intersecting conducting plane forming a " degree if 360/"=n, the number of image charges is n-1.

60

Example 2

Two dielectric media with dielectric constant !1 and !2 are separated by a plane boundary at x=0. A point charge Q exists in medium 1 at distance d from the boundary.

(a) Verify that the field in medium 1 can be obtained from Q and an image charge -Q1 both acting in medium 1.

(b) Verify that the field in medium 2 can be obtained from Q and an image charge +Q2 coinciding with Q, both acting in medium 2.

(c) Determine Q1 and Q2

Q (Point charge)

+Q2 (Image charge)

-Q1 (Image charge)

!1!2

Example 2

Q (Point charge)

+Q2 (Image charge)

-Q1 (Image charge)

!1!2

Example 2

Example 2

Conclusion

! Method of image! Discrete charges

! Conducting bodies of simple boundary

! Method of separation of variables ! Potential of conducting bodies is given.

! Solve differential equations subject to boundary conditions.