1/29/14 Physics for Scientists & Engineers 2, Chapter 23 1

Electric Potential

1/29/14 Physics for Scientists & Engineers 2, Chapter 23 2

Electric Potential Difference ΔV !  The electric potential difference between an initial point i

and final point f is

!  Work and potential are related through

!  The units of electric potential are

!  We express the electric field as

ΔV =Vf −Vi =

U f

q−U i

q= ΔU

q

ΔV = −We

q

1 V = 1 J

1 C

[E]= [F]

[q]= N

C= J/m

C= V

m

1/29/14 Physics for Scientists & Engineers 2, Chapter 23 3

!  If a charged particle moves perpendicular to electric field lines, no work is done

!  If the work done by the electric field is zero, then the electric

potential must be constant !  Equipotential surfaces and lines must always be

perpendicular to the electric field lines !  General observations:

•  The surface of any conductor forms an equipotential surface •  Equipotential surface are always perpendicular to the electric field

lines at any point in space

General Considerations

W = qE ⋅d = 0 if

d ⊥E

1/29/14 Physics for Scientists & Engineers 2, Chapter 23 4

Single Point Charge !  The electric field lines for point charges are radial !  The equipotential surfaces are concentric spheres in 3D !  The equipotential lines take the form of concentric circles in 2D

Positive charge Negative charge

1/29/14 Physics for Scientists & Engineers 2, Chapter 23 5

Electric Potential of Charge Distributions

!  The electric potential is defined as the work required to place a unit charge at a point

!  Work is a force acting over a distance !  The electric field is defined as the force acting on a unit

charge at a point !  So the electric potential is related to the electric field

•  We can determine one given the other

!  To determine the electric potential from the electric field we start with the work done on a charged particle by a force over a displacement

dW =F ⋅ds = q

E ⋅ds

1/29/14 Physics for Scientists & Engineers 2, Chapter 23 6

Electric Potential of Charge Distributions

!  Now we integrate the work as the charge moves in the field

!  Now we can relate the work done to the change in electric potential

!  Taking the standard convention, we can write the electric potential in terms of the electric field

!  In other words, the electric potential is the integral of the electric field

W = q

E ⋅ds

i

f

∫ = qE ⋅ds

i

f

∫

ΔV =Vf −Vi = −We

q= −

E ⋅ds

i

f

∫

V r( )−V ∞( ) =V r( ) = −

E ⋅ds

∞

r

∫

1/29/14 Physics for Scientists & Engineers 2, Chapter 23 7

Electric Potential for a Point Charge

!  Let’s calculate the electric potential due to a point charge !  The magnitude of the electric field is given by

!  The direction of the electric field is radial from the charge !  Integrate along a radial line from ∞ to R

!  So the electric potential due to a point charge is

E r( ) = kq

r 2

V R( ) = −

E ⋅ds

∞

R

∫ = − kqr 2 dr

∞

R

∫ = kqr

⎡⎣⎢

⎤⎦⎥∞

R

= kqR

V = kq

R

1/29/14 Physics for Scientists & Engineers 2, Chapter 23 8

Electric Potential for a Point Charge !  A positive point charge produces a positive potential !  A negative point charge produces a negative potential !  We can calculate the electric potential for all points in an

xy-plane

V x , y( ) = kq

x 2 + y 2

January 29, 2014 Physics for Scientists&Engineers 2 9

Example - Charge moves in E field !  Given is a uniform electric field E. Find

the potential difference Vf-Vi by moving a test charge q0 along the path icf.

!  Idea: Integrate along the path connecting ic and cf. Imagine, we move a test charge q0 from i to c and from c to f.

i

f

c

January 29, 2014 Physics for Scientists&Engineers 2 10

Example - Charge moves in E field !  Given is a uniform electric field E. Find

the potential difference Vf-Vi by moving a test charge q0 along the path icf.

!  Along ic, ds and E are perpendicular

i

f

c

With the distance from c to f: d/sin(45o)

Fixed and Moving Positive Charges PROBLEM:

!  A positive charge of 4.50 μC is fixed in place. !  A particle of mass 6.00 g and charge +3.00 μC is fired with

an initial speed of 66.9 m/s directly toward the fixed charge from a distance of 4.20 cm.

!  How close does the moving charge get to the fixed charge?

SOLUTION: Think

!  The moving charge will gain potential energy as it near the fixed charge.

!  The negative of the change in potential energy of the moving charge is equal to the change in kinetic energy of the moving charge because ΔK + ΔE = 0.

1/29/14 Chapter 23 13

Fixed and Moving Positive Charges Sketch

Research

!  The moving charge gains electric potential energy and loses kinetic energy until it stops.

!  At that point, all the original kinetic energy has been converted into electric potential energy.

1/29/14 Chapter 23 14

Fixed and Moving Positive Charges !  We can express this energy conservation as:

!  The electric potential’s effect on the moving charge is due to the fixed charge:

Simplify

!  Substitute in our expression for the potential difference:

1/29/14 Chapter 23 15

ΔK +ΔU = 0 ⇒ ΔK =−ΔU

0− 12

mv02 =−qmovingΔV ⇒ 1

2mv0

2 = qmovingΔV

ΔV =Vf −Vi = k qfixed

df

−k qfixed

di

= kqfixed1df

−1di

⎛

⎝⎜⎜⎜

⎞

⎠⎟⎟⎟⎟

12

mv02 = qmovingΔV = kqmovingqfixed

1df

−1di

⎛

⎝⎜⎜⎜

⎞

⎠⎟⎟⎟⎟ ⇒

1df

=1di

+mv0

2

2kqmovingqfixed

Fixed and Moving Positive Charges Calculate

!  Putting in our numerical values we get:

Round

!  We round our result to three significant figures:

Double-check !  The final distance of 0.761 cm is less than 4.20 cm.

1/29/14 Chapter 23 16

1df

=1

0.0420 m+

0.00600 kg( ) 66.0 m/s( )2

2 8.99 ⋅109 N m2 /C2( ) 3.00 ⋅10−6 C( ) 4.50 ⋅10−6 C( )1df

=131.485 m−1 ⇒ df = 0.00760545 m

df = 0.00761 m = 0.761 cm

Fixed and Moving Positive Charges !  At the final distance, the electric potential energy of the

moving charge is:

!  At the initial distance, the electric potential energy of the moving charge is:

1/29/14 Chapter 23 17

U = qmovingV = qmoving kqfixed

df

⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟⎟= k

qmovingqfixed

df

U = 8.99 ⋅109 N m2

C2

⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟

3.00 ⋅10−6 C( ) 4.50 ⋅10−6 C( )0.00761 m

=16.0 J

U = qmovingV = qmoving kqfixed

di

⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟⎟= k

qmovingqfixed

di

U = 8.99 ⋅109 N m2

C2

⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟

3.00 ⋅10−6 C( ) 4.50 ⋅10−6 C( )0.0420 m

= 2.90 J

Fixed and Moving Positive Charges !  The initial kinetic energy is:

!  We can see the energy conservation is satisfied:

!  We are confident in our result.

1/29/14 Chapter 23 18

K =

12

mv2 =0.00600 kg( ) 66.0 m/s( )2

2=13.1 J

K =ΔU13.1 J =16.0 J−2.90 J =13.1 J

Finding the Field from the Potential !  We can calculate the electric field from the electric potential

starting with: !  If we look at the component of the electric field along the

direction of ds, we can write the magnitude of the electric field as the partial derivative along the direction s:

!  We can write the components of the electric field in terms of partial derivatives of the potential:

1/29/14 Chapter 23 19

ES =−

∂V∂s

in the direction of ds

−qdV = qE i ds ⇒

E i ds =−dV

Ex =−∂V∂x

; Ey =−∂V∂ y

; Ez =−∂V∂z

E =−

∇V ≡− ∂V

∂x,∂V∂ y

,∂V∂z

⎛

⎝⎜⎜⎜

⎞

⎠⎟⎟⎟⎟ ∇ is the gradient

Graphical Extraction of the Electric Field !  Calculate the magnitude of the electric field at point P. !  To perform this task, we draw a line through point P

perpendicular to the equipotential line reaching from the line of +2000 V to the line of 0 V.

!  The length of this line is 1.5 cm.

!  So the magnitude of the electric field can be approximated as

1/29/14 Chapter 23 20

ES = −

ΔVΔs

Graphical Extraction of the Electric Field !  Putting in our numbers give us:

!  The direction of the electric field points from the positive equipotential line to the negative potential line.

1/29/14 Chapter 23 21

ES =+2000 V( )− 0 V( )

1.5 cm

ES =1.3⋅105 V/m

1/29/14 Physics for Scientists & Engineers 2, Chapter 23 23

System of Point Charges

!  We calculate the electric potential from a system of n point charges by adding the potential functions from each charge

!  This summation produces an electric potential at all points

in space – a scalar function !  Calculating the electric potential from a group of point

charges is usually much simpler than calculating the electric field •  It’s a scalar

V = Vi

i=1

n

∑ = kqi

rii=1

n

∑

Superposition of Electric Potential !  Assume we have a system of three point charges:

q1 = +1.50 μC q2 = +2.50 μC q3 = -3.50 μC

!  q1 is located at (0,a) q2 is located at (0,0) q3 is located at (b,0) a = 8.00 m and b = 6.00 m.

!  What is the electric potential at point P located at (b,a)?

1/29/14 Chapter 23 24

Superposition of Electric Potential !  The electric potential at point P is given by the sum of the

electric potential from the three charges:

1/29/14 Chapter 23 25

V =kqi

rii=1

3

∑

V = kq1

r1

+q2

r2

+q3

r3

⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟⎟= k

q1

b+

q2

a2 + b2+

q3

a

⎛

⎝⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟

V = 8.99 ⋅109 N m2

C2

⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟

1.50 ⋅10−6 C6.00 m

+2.50 ⋅10−6 C

8.00 m( )2+ 6.00 m( )2

+−3.50 ⋅10−6 C

8.00 m

⎛

⎝

⎜⎜⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟⎟⎟⎟⎟

V = 562 V

Electric Potential Energy for a System of Particles

!  So far, we have discussed the electric potential energy of a point charge in a fixed electric field.

!  Now we introduce the concept of the electric potential energy of a system of point charges.

!  In the case of a fixed electric field, the point charge itself did not affect the electric field that did work on the charge.

!  Now we consider a system of point charges that produce the electric potential themselves.

!  We begin with a system of charges that are infinitely far apart, U = 0, by convention.

!  To bring these charges into proximity with each other, we must do work on the charges, which changes the electric potential energy of the system.

1/29/14 Chapter 23 26

Two Charges !  To illustrate the concept of the electric potential energy of a

system of particles we calculate the electric potential energy of a system of two point charges, q1 and q2.

!  We start our calculation with the two charges at infinity. !  We then bring in point charge q1. Because there is no

electric field and no corresponding electric force, this action requires no work to be done on the charge.

!  Keeping this charge (q1) stationary, we bring the second point charge (q2) in from infinity to a distance r from q1.

!  This requires work q2V1(r).

1/29/14 Chapter 23 27

!  The electric potential energy of this two charge system is:

!  The electric potential of the two charge system is:

!  If the two point charges have the same sign, then we must

do positive work on the particles to bring them together from infinity (i.e., we must put energy into the system).

!  If the two charges have opposite signs, we must do negative work on the system to bring them together from infinity (i.e., energy is released from the system).

Two Charges

U = q2V1(r) where V1(r) =

kq1

r

1/29/14 Chapter 23 28

U =

kq1q2

r

Three Charges !  Now let’s try three charges. !  Consider three point charges, q1, q2,

and q3, at fixed positions arranged an equal distance d from each other.

!  What is the electric potential energy U of the assembly of these charges?

!  The potential energy is equal to the work we must do to assemble the system, bringing in each charge from an infinite distance.

!  Let’s build the system by bringing the charges in from infinity, one at a time.

1/29/14 Chapter 23 29

© W. Bauer and G.D. Westfall

Three Charges !  Bringing in q1 requires no work. !  With q1 in place, bring in q2:

!  Now bring in q3:

!  The total electric potential of the system is then:

1/29/14 Chapter 23 30

U12 =W = q2V = q2k

q1

d

W13 +W23 =U13 +U23 = q3 V1 +V2( )

U13 +U23 = q3 kq1

d+ k

q2

d⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟

U123 =U12 +U13 +U23

U123 = q2kq1

d+ q3 k

q1

d+ k

q2

d⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟=

kd

q1q2 + q1q3 + q2q3( )

© W. Bauer and G.D. Westfall