Today 3/12

Plates if charge E-Field Potential

HW: “Plate Potential” Due Friday, 3/14

(+)

How big is E?(-)

Enet = 0

02E

02

E 0

netE

Enet = 0

0 = Q/A

Charged conducting plate

What’s wrong with this picture?

A = Area of one side

L = 0/2

Charged conducting plate

Free charge always goes to surface of conductor.

R = 0/2

L = 0/2

Charged conducting plate

R = 0/2

What is the electric field inside the conductor?

EL = L/20 = 0/40

L = 0/2

Charged conducting plate

R = 0/2

What is the electric field inside the conductor?

ER = R/20 = 0/40

L = 0/2

Charged conducting plate

R = 0/2

What is the electric field inside the conductor?

The electric field is zero everywhere inside the conductor.Always, any conductor, no exceptions.

q=+1C

What happens if I let it go?

KE = 0 KE = 100 J

A B

Assume the particle gains 100 joules of kinetic energy as it moves from A to B.

q=+1C

A B

Now I stop it at B.

How much work must I do to move it back to A? +100 J

How does the potential energy change in moving from B to A? +100 J PEBA = +100 J

PEAB = -100 J

q=+1C

A B

PEBA = +200 JPEAB = -200 J

What if q=+2C?

How much work must I do to move it back to A?

How does the potential energy change in moving from B to A?

+200 J

+200 J

q=+2C

A B

Now we are back to our original definition.

PEBA = (+100 J/C)x(q)

VBA tells us how much PE changes when +1C is moved from B to A.

PEBA = VBA q

VBA = +100 J/C

q=+1C

A B

What if q = -1C?

First I must turn my hand around.

q= -1C

A B

What if q= -1C?

How much work must I do to move it back to A? -100 J

How does the potential energy change in moving from B to A? -100 J

PEBA = (+100 J/C)x(q)

PEBA = (+100 J/C)x(-1)

VBA tells us how much PE changes when +1C is moved from B to A.

PEBA = VBA q

q= -1C

A B

What if q= -1C?

How much work must I do to move it back to A? -100 J

How does the potential energy change in moving from B to A? -100 J

PEBA = (+100 J/C)x(q)

PEBA = (+100 J/C)x(-1)

VBA does not depend on the sign of the point charge but PEBA does!!!!!

PEBA = VBA q

q= -1C

A B

VAB = -100 J/C

VAB = -100 volts

A proton is released from rest at A. What is its speed when it reaches B?

m= 1.7 x 10-27 kg

q= 1.6 x 10-19 C

PEAB = q VAB

PEAB = q -100 J/C)

PEAB = -1.6 x 10-17 JWhat happens to the kinetic energy?

A B

KEAB = +1.6 x 10-17 J

A proton is released from rest at A. What is its speed when it reaches B?

m= 1.7 x 10-27 kg

q= 1.6 x 10-19 C

mv2 = +1.6 x 10-17 J

v = 1.4 x 105 m/sWhat happens to the kinetic energy?

PEAB = -1.6 x 10-17 J

A BA B

What direction is the force on an electron?

EF

A B

VAB = -100 J/C

VBA = +100 J/C

An electron is released from rest at B. What is its speed when it reaches A?

m= 9.1 x 10-31 kg

q= -1.6 x 10-19 C PEBA = q VBA

PEBA = q +100 J/C)

PEBA = -1.6 x 10-17 JWhat happens to the kinetic energy?

A B

KEBA = +1.6 x 10-17 J

mv2 = +1.6 x 10-17 J

v = 5.9 x 106 m/s

An electron is released from rest at B. What is its speed when it reaches A?

m= 9.1 x 10-31 kg

q= -1.6 x 10-19 C

What happens to the kinetic energy?

PEBA = -1.6 x 10-17 J

B A

VAB = ?

B A

How does doubling the E-field affect VAB ?

VAB = ?

VAB

doubles

Bnew A

How does moving point B affect VAB ?

VAB = ?

How does VAB

depend on E and D?

D

Anything else?

VAB is halved

Bold

B A

KEAB = 0

constantspeed

v0v0

KEA=1/2 mv02

KEB=1/2 mv02

How much work must I do to move the charge from A to B? D

B A

How much work must I do to move the charge from A to B? FE

= qEFHq

WAB = FE x D

WAB = qED

WAB = FHq x D

B A

What is the change in potential energy in going from A to B?

FE = qEFHq

WAB = qED

PEAB = qED

PEAB = qVAB

VAB = EDAB

WAB = qED

Only applies when the field is uniform over the distance. VAB‘s sign depends on the direction of E. In this case it’s positive.