Electric Potential

Chapter 25 The Electric Potential Equipotential Surfaces Potential Due to a Distribution of Charges Calculating the Electric Field From the Potential

ELECTRICAL POTENTIAL DIFFERENCE

Electrical Potential = Potential Energy per Unit Charge 

V AB =



U AB / q



V AB =



U AB / q = - (1/q)



B q E . dL = A



A B E . dL



V AB = -



A B E . dL

Remember: dL points from A to B 

V AB =

Electrical potential difference between the points A and B

ELECTRICAL POTENTIAL IN A CONSTANT FIELD E E

A • dL L • B 

V AB =



U AB / 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 

V AB = V B – V A = -W AB /q = -



E.dl

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

V AB = -



E.dl =



E dl = E



dl = E L



V AB = E L



U AB = q E L

Example: Electric potential of a uniform electric field



V



V b



V a

 

a b



E dx

 

b



a E

  

Ed dr

A positive charge would be pushed from regions of high potential to regions of low potential.

a d d

r

b a d d

r

b E E Remember: since the electric force is conservative, the potential difference does not depend on the integration path, but on the initial and final points.

The Electric Potential Point Charge q

 What is the electrical potential difference between two points (a and b) in the electric field produced by a point charge q.

q b a

F

=q t

E

The Electric Potential

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

b c First find the work done by q’s field when q t 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 W(c to b) =

F



dr

r c r b  q t E (r )  dr  r a r b  q t E(r)dr q a  kq t q r a r b  dr r 2

F

=q t

E

The Electric Potential

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

b c First find the work done by q’s field when q t 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 W(c to b) = hence

W

 r c r b  q t E (r )  kq t q   1 r a  dr  1  r a r b  r b  q t

F



dr

E(r)dr  q a kq t q r a r b  dr r 2

F

=q t

E

The Electric Potential

U(r b )  U(r a )   W  kq t q  1  r b  1 r a  And since 

V AB =



U AB / q t

q c b a

F

=q t

E



V AB = k q [ 1/r b – 1/r a ]

The Electric Potential



V AB = k q [ 1/r b – 1/r a ]

b From this it’s natural to choose the

zero of electric potential

to be when r a  c Letting

a

be the point at infinity, and dropping the subscript

b

, we get the electric potential: V = k q / r q When the source charge is

q

, and the electric potential is evaluated at the point

r

.

a Remember: this is the electric potential with respect to infinity

F

=q t

E

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 …

Potential Produced by a Continuous 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: dq i



Potential Produced by a Continuous 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 dV A = k dq / r

 Remember: k=1/(4  0 )

dq V A =



vol dV A =



k dq / r vol

Example: a disk of charge

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. P dq = s 2  wdw

dV



V

  1 4  0 

dV dq r

  s 2  0 1 4  0 

R

0 (

w

2 s  2 

wdw w

2 

z

2 )  1 2

z

2

wdw

dw z r

V

 s 2  0 (

R

2 

z

2 

z

) w R

A Equipotential Surfaces (lines) E B

Since the field E is constant  V AB = E L

X L E

Then, at a distance X from plate A  V AX = E X All the points along the dashed line, at X, are at the same potential.

The dashed line is an equipotential line L

X Equipotential Surfaces (lines) E

It takes no work to move a charge at right angles to an electric field E  dL   E•dL = 0   V = 0

L

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



ELECTRIC FIELD

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 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 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: Lines of constant V (perpendicular to E) Lines of constant E

Equipotential plots are like contour maps of hills and valleys.

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.

Force and Potential Energy

Choosing an arbitrary reference point

r

0 (such as at which U(

r

0 ) = 0, the potential energy is:  )

U

(

x

,

y

,

z

)  

r 0 r



F



dr F

(

x

,

y

,

z

)   (ˆ  

x

 This can be inverted:  

y

  

z

)

U

(

x

,

y

,

z

) The potential energy

U

is calculated from the force

F

, and conversely the force

F

can be calculated from the potential energy

U

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

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. P dq = s 2  wdw

dV



V

  1 4  0 

dV dq r

  s 2  0 1 4  0 

R

0 (

w

2 s  2 

wdw w

2 

z

2 )  1 2

z

2

wdw V

 s 2  0 (

R

2 

z

2 

z

) dw z w r R

Example: a disk of charge

V

(

z

)  s 2  0 (

R

2 

z

2 

z

) By symmetry one sees that E x =E y =0 at P.

Find E z from

E z

  

V



z

  s 2  0 

R

2

z



z

2  1   P z r

This is easier than integrating over the components of vectors. Here we integrate over a scalar and then take partial derivatives.

dw w R

Example: point charge

Put a point charge q at the origin.

Find V(

r

): here this is easy:

V

(

r

) 

k q r

q

r

Example: point charge

Put a point charge q at the origin.

Find V(

r

): here this is easy:

V

(

r

)  Then find

E

(

r

) from the derivatives: E (

r

)   (ˆ   x  ˆ j   y 

k q r

 q

r

 z)V(x, y, z)

Example: point charge

Put a point charge q at the origin.

Find V(

r

): here this is easy:

V

(

r

) 

k q r

q

r

Then find

E

(

r

) from the derivatives: E (

r

)   (ˆ   x  ˆ j   y  Derivative:  

x

1

r

  

x x

2  1

y

2 

z

2     z)V(x, y, z) 1 2 

x

2 

y

2 2

x



z

2  3 2

Example: point charge

Put a point charge q at the origin.

Find V(

r

): here this is easy:

V

(

r

) 

k q r

q

r

Then find

E

(

r

) from the derivatives: E (

r

)   (ˆ   x  ˆ j   y  Derivative:  

x

1

r

  

x x

2  1

y

2 

z

2     z)V(x, y, z) 1 2 

x

2 

y

2 2

x



z

2  3 2 So:

E

(

r

) 

kq x

ˆ 

y

ˆ 

z

ˆ 

r

3

r

kq r

3 

kq r

2