Prof. David R. JacksonECE Dept.

Fall 2014

Notes 4

ECE 2317Applied Electricity and Magnetism

Notes prepared by the EM Group University of Houston

1

so

Electric Field

F q E

1E F

q

The electric-field vector is the force vector for a unit charge.

Units of electric field: [V/m]

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

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

+-

V0 [V] Eq

Note: The “test” charge is small enough so that it does not disturb the charge on the capacitor and hence the field from the capacitor.

2

Electric Field (cont.)

20

ˆ4

qE r

r

Note: The electric-field vector may be non-uniform.

E

qPoint charge q in free space:

3

Voltage Drop

The voltage drop between two points is now defined.

Volta

4

B

AB

A

V V A V B E dr

CA

B

E (x,y,z)

Comment: In statics, the line integral is independent of the shape of the path.

(This will be proven later after we talk about the curl.)

Voltage Drop (Cont.)

EF q E

We now explore the physical interpretation of voltage drop.

CA

B

E (x,y,z)

q test charge

F ext (x,y,z)

A test charge is moved from point A to point B at a constant speed (no increase in kinetic energy).

ext EF F q E 5

Voltage Drop (cont.)

W ext = work done by observer in moving the charge.

B B Bext ext

A A A

W F dr q E dr q E dr

6

CA

B

E (x,y,z)

q test charge

F ext (x,y,z)

extABW qVSo

Voltage Drop (cont.)

so

extABW qV

extW PE B PE A But we also have

ABqV PE B PE A

or

1ABV PE A PE B

q

7

Physical Interpretation of Voltage

8

The voltage drop is equal to the change in potential energy of a unit test charge.

Point A is at a higher voltage, and hence a higher potential energy, than point B.

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

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

+-

V0 [V] Eq = 1A

B

1ABV PE A PE B q

Comments

9

The electric field vector points from positive charges to negative charges.

Positive charges are at a higher voltage, and negative charges are at a lower voltage.

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

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

+-

V0 [V] E

Review of Doing Line Integrals

10

B

AB

A

V E dr In rectangular coordinates,

ˆ ˆ ˆx y zE x E y E z E

ˆ ˆ ˆ ˆˆ ˆr x x y y z z dr x dx y dy z dz

B

AB x y z

A

V E dx E dy E dz Hence

Note: The limits are vector points.

, , , , , ,B B B

A A A

x y z

AB x y z

x y z

V E x y z dx E x y z dy E x y z dz so

Each integrand must be parameterized in terms of the respective integration variable. This requires knowledge of the path C.

Example

Find: E

Assume: 0ˆ, ,E x y z x E

0 V, , [ ]B

AB

A

V E x y z dr V

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

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

+-

V0 [V] E

x = 0

x = h

A

B

11

Ideal parallel-plate capacitor assumption.

Example (cont.)

Evaluate in rectangular coordinates:

0 V, , [ ]B

AB

A

V E x y z dr V

ˆ ˆ ˆx y zE x E y E z E

ˆ ˆ ˆdr x dx y dy z dz

x y zE dr E dx E dy E dz

0 V[ ]B

A

x

AB x

x

V E dx V 12

Example (cont.)

Hence, we have

0

B

A

x

AB x

x

V E dx V

0 0

0

h

E dx V

0 0E h V 0 0 /E V h

0ˆ, ,E x y z x E

0 V/mˆ, , [ ]V

E x y z xh

Recall that

13

Example

A proton is released at point A on the top plate with zero velocity. Find the velocity v(x) of the proton at distance x from the top plate (at point B).

0 0KE x KE PE PE x

210 0

2mv x PE PE x q V V x

14

Conservation of energy:

h = 0.1 [m]

V0 = 9 [V]

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

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

+-V0 [V] E

x = 0

x = h

ProtonA

B x = x

x = 0

Example (cont.)

210

2mv x q V V x

0 0

0 0

0B x x

x

A

V VV V x E dr E dx dx x

h h

Hence: 2 01

2

Vmv x q x

h

02qVv x x

h m

15

Example (cont.)

Hence:

02qVv x x

h m

h = 0.1 [m]V0 = 9 [V]

q = 1.602 x 10-19 [C]

m = 1.673 x 10-27 [kg]

6 m/s5.627 10 [ ]v x x

(See Appendix B of the Hayt & Buck book or Appendix D of the Shen & Kong book.)

16

Reference Point A reference point R is a point where the voltage is assigned.

This makes the voltage unique at all points.

The voltage at a given point is often called the potential .

R 0R

C

r = (x,y,z)

Note: 0 is often chosen to be zero, but this is not necessary.

17

Integrating the electric field along C will determine the potential at point r.

Example

Find the potential on the left terminal (cathode) assuming R is on right terminal (anode) and = 0 at the reference point.

9BAV B A

V9 [ ]A

18

9 [V]

A B 0R

9 [V]

+-

9BAV

A B

Find the potential function at x, assuming that the reference point R is on the bottom plate, and the voltage at R is zero.

Example

19

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

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

+-

V0 [V] E

x = 0

x = h

r (x, y, z)

0R

0 0

0 0

0x x

x

V Vx E dx dx x

h h

00 h V

00 V[ ]

Vx V x

h

Also,

Hence