electric potential chapter 25 electric potential energy electric potential equipotential surfaces

Electric Potential

Chapter 25Electric Potential Energy

Electric PotentialEquipotential Surfaces

Electric Potential: the Bottom Line

The electric potential V(r) is an easy way to calculate the electric field – easier than directly using Coulomb’s law.

For a collection of charges qi at positions ri the electric potential at r is the scalar sum:

€

V (r r ) = 1

4πε0

qr

€

V (r r ) = 1

4πε0

qir r −

r r ii

∑

For a point charge at the origin:

Calculate V and then find the electric field by taking the gradient:

€

r E (

r r ) = −

r ∇V (

r r ) = − ∂V

∂xˆ i + ∂V

∂yˆ j + ∂V

∂zˆ k

⎛ ⎝ ⎜

⎞ ⎠ ⎟

Example: two point charges

€

V (x, y) = 14πε0

qx 2 + y 2

+ q

x − a( )2 + y 2

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟

€

Ex (x,y) = −∂V∂x

= − q4πε0

− 12( )

2x

x 2 + y 2( )3

2+

⎛

⎝

⎜ ⎜

+ − 12( )

2 x − a( )

x − a( )2 + y 2

( )3

2

⎞

⎠

⎟ ⎟ ⎟

aq q

r

CONSERVATIVE FORCESA conservative force “gives back” work that has been done against it

When the total work done by a force on an object moving around a closed loop is zero, then the force is conservative

F·dr = 0 F is conservative

The circle on the integral sign indicates that the integral is taken over a closed path

The work done by a conservative force, in moving and objectbetween two points A and B, is independent of the path taken

W=AB F·dr is a function of A and B only – it is

NOT a function of the path selected. We can define a potential energy difference as UAB=-W.

°

POTENTIAL ENERGY

Potential energy is defined at each point in space, but it is only the difference in potential energy that matters.

Potential energy is measured with respect to a reference point (usually infinity). So we let A be the reference point (i.e, define UA to be zero), and use the above integral as the definition of U at point B.

The change UAB in potential energy due to the electric force is thus

UAB = -q AB E·dr

UAB = UB – UA = potential energy difference between A and B

POTENTIAL ENERGY IN A CONSTANT FIELD E

E

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

UAB = UB – UA = - q E·dl

But E = constant, and E.dl = -E dl, so:

UAB = - q E · dl = q E dl = q E dl = q E L

UAB = q E L

• •A BL

dl

ELECTRIC POTENTIAL DIFFERENCE

The potential energy U depends on the charge being moved.To remove this dependence, we introduce the concept of the electric potential V. This is defined in terms of the difference V:

VAB = UAB / q = - AB E · dl

Electrical Potential = Potential Energy per Unit Charge = line integral of -E·dl

VAB = Electric potential difference between the points A and B. Units are Volts (1V = 1 J/C), and so the electric potential is often called the voltage.A positive charge is pushed from regions of high potential to regions of low potential.

ELECTRIC POTENTIAL IN A CONSTANT FIELD E

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 = - E·dl

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

VAB = - E·dl = E dl = E dl = E L

VAB = E L

• •A BL

E

dL VAB = UAB / q

UAB = q E L

Cases in which the electric field E is not aligned with dl

VAB = - AB E · dl•

A

B

E

•

E . dl = E dl cos VAB = - E cos dl = - E L cos

The electric potential difference does not depend on the integration path. So pick a simple path.

One possibility is to integrate along the straight line AB.This is convenient in this case because the field E is constant, and the angle between E and dl is constant.

A

B

Cases in which the electric field E is not aligned with dl

• •A

B

E

•

Cd

Another possibility is to choose a path that goes from A to C, and then from C to B

VAB = VAC + VCB VAC = E d VCB = 0 (E dL)

Thus, VAB = E d but d = L cos = - L cos

VAB = - E L cos

L

VAB = - AB E · dl

Equipotential Surfaces (lines)

VAB = E L

For a constant field E

E

L

E

L

BA

x

VAx = E x

All the points along the dashed lineat x, are at the same potential.

The dashed line is an equipotential line

Similarly, at a distance x from plate A

Equipotential Surfaces (lines)

E

L

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 ELECTRIC FIELD

We can make graphical representations of the electric potential in the same way as we have created for the electric field:

Equipotential Surfaces

Lines of constant E



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



Equipotential plots are like contour maps of hills and valleys. A positive charge would be pushed from hills to valleys.

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


Equipotential plots are like contour maps of hills and valleys.

How do the equipotential surfaces look for:(a) A point charge?

(b) An electric dipole?

+

+ -

E

Equipotential plots are like contour maps of hills and valleys. A positive charge would be pushed from hills to valleys.

Point Charge q

The Electric Potential

b

aq

What is the electrical potential differencebetween two points (a and b) in the electricfield produced by a point charge q.

Electric Potential of a Point Charge

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


c

b

aq

Choose a path a-c-b.

Vab = Vac + Vcb Vab = 0 because on this path

r F ⊥d

r r

q t

r E (

r r )• d

r r

r r c

r r b

= tra

rb

E(r)dr=ktdrr2ra

rb

Vbc =


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


F=qtE

c

b

aq

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

r F ⊥d

r r

=kq tq1ra

−1rb

⎛ ⎝ ⎜ ⎞

⎠ ⎟

q t

r E (

r r )• d

r r

r r c

r r b

= tra

rb

E(r)dr=ktdrr2ra

rb

hence W

W(c to b) =



U(

r r b ) −U(

r r a) =−W =kt

1rb−

1ra

⎛ ⎝ ⎜ ⎞

⎠ ⎟

F=qtE

c

b

aqVAB = UAB / qtAnd since

VAB = k q [ 1/rb – 1/ra ]



From this it’s natural to choose the zero of electric potential

to be when ra

Letting a be the point at infinity, and droppingthe subscript b, we get the electric potential:

When the source charge is q,and the electric potential isevaluated at the point r.

F=qtE

c

b

aq

VAB = k q [ 1/rb – 1/ra ]

V = k q / r

Remember: this is the electric potential with respect to infinity


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 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:

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:

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

dq

A

dVA = k dq / rr

VA = dVA = k dq / rvol vol

Remember:k=1/(40)

Example: a disk of chargeSuppose the disk has radius R and a charge per unit area .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 thepotential 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.

dw

P

r

Rw

z

dq = 2wdw

dV =1

40

dr=

14

0

2wdww2 + z2

∴V = dV =2

0

(w2

0

R

+ z2 )−12wdw

V =2

0

( R 2 + z2 −z)

electric potential chapter 25 electric potential energy electric potential equipotential surfaces

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