Potential near a point charge

A

B

+

Potential near a point charge

What is the potential difference between A and B?

1q

0,0,

0,0,

dxsd

EE x

+1q

A

Bsd

E

x

Now integrate along the path:

B

A

x

x

B

A

x

B

A

AB dxx

qdxEdVV

21

04

1

dxEdV

dxEsdEdV

x

x

0,0,0,0,

Change in potential along a short section of the path:

0,0,

0,0,

dxsd

EE x

+1q

A

Bsd

E

x

dxEdV

dxEsdEdV

x

x

0,0,0,0,

Change in potential along a short section of the path:

ABAB xx

qV11

4

11

0

Potential difference near a point charge

ABAB rr

qV11

4

11

0

+1q

AB

Ar

Br

Potential at one location

11

4

11

0 BAB r

qV

+1q

B

Let rA go to infinity…

Potential at one location

r

qrV 1

04

1)(

1q+

r

The potential at a distance r from a point charge, relative to infinity:

r

qrV 1

04

1)(

Potential at one location

r

qrV 1

04

1)(

1q

The potential at a distance r from a point charge, relative to infinity:

+r

2q

Potential energy of two charges

r

qqrU 21

04

1)(

1q

The potential energy of two point charges, relative to infinity:

+r

+2q

Potential energy of a system of charges

01 U

12

21

02 4

1

r

qqU

23

32

013

31

03 4

1

4

1

r

qq

r

qqU

23

32

13

31

12

21

0tot 4

1

r

qq

r

qq

r

qqU

Finding the field from the potential

dzEdyEdxE

sdEdV

zyx

The change in potential along a very small path:

sd

E

E

E

Finding the field from the potential

dzEdyEdxEdV zyx

Choose a path that only goes in the x-direction (dy = dz = 0):

sd

E

E

E

Finding the field from the potential

dxEdV x

Choose a path that only goes in the x-direction (dy = dz = 0):

sd

E

E

E

dx

dVEx (holding y and z

fixed)

Finding the field from the potential

dxEdV x

Choose a path that only goes in the x-direction (dy = dz = 0):

sd

E

E

E

x

VEx

(partial derivative)

Finding the field from the potential

dxEdV y

Choose a path that only goes in the y-direction (dx = dz = 0):

sd

E

E

E

y

VEy

Finding the field from the potential

dzEdV z

Choose a path that only goes in the z-direction (dx = dy = 0):

sd

E

E

E

z

VEz

Electric field is the negative gradient (梯度 ) of the

potential

z

VE

y

VE

x

VE zyx

,,

V

xxE xE

Electric field is the negative gradient (梯度 ) of the

potential

zyxVE

,, where,

V

xxE xE

The potential is like the height of the hill.

The field is like the slope of the hill.

Just remember:- positive charges go down the hill- negative charges go up!

Field around a point charge

r

qrV 1

04

1)(

1q

The potential near a point charge, relative to infinity:

+r

r

q

dr

d

dr

dVrE 1

04

1)(

The field strength is the gradient of the potential:

Field around a point chargeThe potential near a point charge, relative to infinity:

21

04

1)(

r

qrE

r

qrV 1

04

1)(

1q+

r

The field strength is the gradient of the potential:

Potential along the axis of a ring

Potential obeys the superposition principle, just like the field.

R

x

22 Rx

2204

1

Rx

dQdV

Potential due to one small piece:

dQ

Potential along the axis of a ring

Potential obeys the superposition principle, just like the field.

R

x

22 Rx

dQ

RxdVV

220

1

4

1

Integrate:

dQ

Potential along the axis of a ring

Potential obeys the superposition principle, just like the field.

R

x

22 Rx

2204

1

Rx

QV

Integrate:

dQ

Field along the axis of a ring

The strength of the field is the negative of the potential gradient:

220

1

4 Rxx

Q

x

VEx

E

Field along the axis of a ring

The strength of the field is the negative of the potential gradient:

2/32204

1

Rx

QxEx

E

Field along the axis of a ringWe already calculated this field the hard way.

It is often easier to first calculate the potential, then use its gradient to get the field.

2/32204

1

Rx

QxEx

E

Potential due to a uniformly charged sphere

+ + +

++

+

++

++

++

Q

Remember: The field outside a charged sphere is the same as the field of a point charge.

The same is true for the potential.

Potential due to a uniformly charged sphere

+ + +

++

+

++

++

++

Q r

r

QrV

04

1)(

V(∞) = 0

Potential at the surface

+ + +

++

+

++

++

++

Q

R

R

QV

0surface 4

1

V(∞) = 0

Potential at the surface

+ + +

++

+

++

++

++

Q

R

0surface

RV

Define the surface charge density:

24 R

Q

A

Q

V(∞) = 0

Surface potential of a protein

Positive (+)Negative (-)

0surface

RV

Atserpin 1

Arabidopsis thaliana

Potential in a conductor

+

At equilibrium, the field inside the conductor must be zero.

A

B

0AB V

Potential in a conductor

+

So the potential inside a conductor (and at the surface) must be constant.

constantV

Example: A negatively charged metal sphere