Electric potential
Download
1 / 27

Electric Potential - PowerPoint PPT Presentation


  • 125 Views
  • Updated On :

Electric Potential. Chapter 25 The Electric Potential Equipotential Surfaces Potential Due to a Distribution of Charges Calculating the Electric Field From the Potential.  V AB =  U AB / q. Electrical Potential = Potential Energy per Unit Charge. B. B.

loader
I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.
capcha
Download Presentation

PowerPoint Slideshow about 'Electric Potential' - consuelo


An Image/Link below is provided (as is) to download presentation

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.


- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript
Electric potential l.jpg

Electric Potential

Chapter 25

The Electric Potential

Equipotential Surfaces

Potential Due to a

Distribution of Charges

Calculating the Electric Field

From the Potential


Slide2 l.jpg

VAB = UAB / q

Electrical Potential = Potential Energy per Unit Charge

B

B

VAB = UAB / q = - (1/q)  q E . dL = -  E . dL

VAB = -  E . dL

B

A

A

A

ELECTRICAL POTENTIAL DIFFERENCE

Remember:

dL points

from A to B

VAB = Electrical potential difference between the points A and B


Slide3 l.jpg

E

L

A

B

dL

VAB = E L

ELECTRICAL POTENTIAL IN A CONSTANT FIELD E

VAB = UAB / q

The electrical potential difference between A and B equals the work per unit charge necessary to move a charge +q from A to B

VAB = VB – VA = -WAB /q = -  E.dl

But E = constant, and E.dl = -1 E dl, then:

VAB = -  E.dl =  E dl = E  dl = E L

UAB =q E L


Slide4 l.jpg

dr

E

a

b

d

a

dr

E

b

d

Example: Electric potential of a uniform electric field

A positive charge would be pushed

from regions of high potential

to regions of low potential.

Remember: since the electric force is conservative, the potential difference

does not depend on the integration path, but on the initial and final points.


Point charge q l.jpg

b

F=qtE

a

q

The Electric Potential

What is the electrical potential difference

between two points (a and b) in the electric

field produced by a point charge q.

Point Charge q 


Place the point charge q at the origin the electric field points radially outwards l.jpg

b

c

F=qtE

a

q

The Electric Potential

First find the work done by q’s field when qt is moved from a to b on the path a-c-b.

W = W(a to c) + W(c to b)

W(a to c) = 0 because on this path

Place the point charge q at the origin.

The electric field points radially outwards.

W(c to b) =


Place the point charge q at the origin the electric field points radially outwards7 l.jpg

b

c

F=qtE

a

q

The Electric Potential

First find the work done by q’s field when qt is moved from a to b on the path a-c-b.

W = W(a to c) + W(c to b)

W(a to c) = 0 because on this path

Place the point charge q at the origin.

The electric field points radially outwards.

W(c to b) =

hence W


Slide8 l.jpg

b

c

VAB = UAB / qt

F=qtE

a

q

VAB = k q [ 1/rb – 1/ra ]

The Electric Potential

And since


Slide9 l.jpg

b

c

F=qtE

a

q

VAB = k q [ 1/rb – 1/ra ]

V = k q / r

The Electric Potential

From this it’s natural to choose

the zero of electric potential

to be when ra

Letting a be the point at infinity, and dropping

the subscript b, we get the electric potential:

When the source charge is q,

and the electric potential is

evaluated at the point r.

Remember: this is the electric potential with respect to infinity


Potential due to a group of charges l.jpg
Potential Due to a Group of Charges

  • For isolated point charges just add the potentials created by

  • each charge (superposition)

  • For a continuous distribution of charge …


Slide11 l.jpg

dqi

Potential Produced by aContinuous Distribution of Charge

In the case of a continuous distribution of charge we first

divide the distribution up into small pieces, and then we

sum the contribution, to the electric potential, from each

piece:


Slide12 l.jpg

VA =  dVA =  k dq / r

vol

vol

Potential Produced by aContinuous Distribution of Charge

In the case of a continuous distribution of charge we first

divide the distribution up into small pieces, and then we

sum the contribution, to the electric potential, from each

piece:

In the limit of very small pieces, the sum is an integral

A

r

dVA = k dq / r

Remember:

k=1/(40)

dq


Example a disk of charge l.jpg

P

r

z

dw

w

R

Example: a disk of charge

dq = s2pwdw

Suppose the disk has radius R and a charge per unit area s.

Find the potential at a point P up the z axis (centered on the disk).

Divide the object into small elements of charge and find the

potential dV at P due to each bit. For a disk, a bit (differential

of area) is a small ring of width dw and radius w.


Slide14 l.jpg

A

B

E

E

VAB = E L

VAX = E X

X

L

L

Equipotential Surfaces (lines)

Since the field E is constant

Then, at a distance X from plate A

All the points along the dashed line,

at X, are at the same potential.

The dashed line is an

equipotential line


Slide15 l.jpg

E

L

EQUIPOTENTIAL  ELECTRIC FIELD

Equipotential Surfaces (lines)

X

It takes no work to move a charge

at right angles to an electric field

E dL   E•dL = 0  V = 0

If a surface (line) is perpendicular to

the electric field, all the points in

the surface (line) are at the same

potential. Such surface (line) is called

EQUIPOTENTIAL


Equipotential surfaces l.jpg

Equipotential Surfaces

We can make graphical representations of the

electric potential in the same way as we have

created for the electric field:

Lines of constant E


Equipotential surfaces17 l.jpg

Lines of constant V

(perpendicular to E)

Lines of constant E

Equipotential Surfaces

We can make graphical representations of the

electric potential in the same way as we have

created for the electric field:


Equipotential surfaces18 l.jpg

Lines of constant V

(perpendicular to E)

Lines of constant E

Equipotential plots are like contour maps of hills and valleys.

Equipotential Surfaces

We can make graphical representations of the

electric potential in the same way as we have

created for the electric field:


Equipotential surfaces19 l.jpg

Equipotential Surfaces

How do the equipotential surfaces look for:

(a) A point charge?

E

+

(b) An electric dipole?

-

+

Equipotential plots are like contour maps of hills and valleys.


Slide20 l.jpg

This can be inverted:

The potential energy U is calculated from the force F,

and conversely

the force Fcan be calculated from the potential energy U

Force and Potential Energy

Choosing an arbitrary reference point r0 (such as )

at which U(r0) = 0, the potential energy is:


Field and electric potential l.jpg
Field and Electric Potential

Dividing the preceding expressions by the (test) charge q we obtain:

V (x, y, z) = -  E• dr

E (x, y, z) = - V

V = (dV/dx) i + (dV/dy) j + (dV/dz) k

  gradient


Slide22 l.jpg

P

r

z

dw

w

R

Example: a disk of charge

  • Suppose the disk has radius R and a charge per unit area s.

  • Find the potential and electric field at a point up the z axis.

  • Divide the object into small elements of charge and find the

  • potential dV at P due to each bit. So here let a bit be a small

  • ring of charge width dw and radius w.

dq = s2pwdw


Slide23 l.jpg

P

r

z

dw

w

R

Example: a disk of charge

By symmetry one sees that Ex=Ey=0 at P.

Find Ez from

This is easier than integrating over the

components of vectors. Here we integrate

over a scalar and then take partial derivatives.


Slide24 l.jpg

Example: point charge

Put a point charge q at the origin.

Find V(r): here this is easy:

r

q


Slide25 l.jpg

Example: point charge

Put a point charge q at the origin.

Find V(r): here this is easy:

r

q

Then find E(r) from the derivatives:


Slide26 l.jpg

Example: point charge

Put a point charge q at the origin.

Find V(r): here this is easy:

r

q

Then find E(r) from the derivatives:

Derivative:


Slide27 l.jpg

Example: point charge

Put a point charge q at the origin.

Find V(r): here this is easy:

r

q

Then find E(r) from the derivatives:

Derivative:

So: