Electric Potential

Gravitational Potential Energy

!Fg = G

m1m2

r2r̂ ,

!g =!Fgm

= !GMr2

r̂

Wg =!Fg! "d!s = #GMm

r2$%&

'()

rA

rB

! dr = GMmr

*+,

-./rA

rB

=GMm 1rB# 1rA

$%&

'()

Wg =!Fg! "d!s = mgcos#

A

B

! ds = $ mgcos%A

B

! ds

= $ mgdy = $mg yB $ yA( )A

B

!

!F"! "d!s = 0

Electric Potential Energy & Electric Potential

!U =UB "UA = "W = "!F

A

B

# $d!l = " q

!E

A

B

# $d!l

We =

!Fe! "d!l

!Fe =

14!"0

q1q2r2

r̂

V = Uq0

= 14!"0

qr

#V = $ #Uq0

= $ Ub

q0

$Ua

q0

%&'

()*= $ Vb $Va( ) =Va $Vb

Va $Vb = $!E +d!l

b

a

,

Gravity & Electric Forces Gravity and Electric forces are conservative force and when you deal with conservative forces, the work that has to be done in going from one point to the other is independent of the path. This is the definition of conservative force

Gravitational force Electric force

Electric Potential

!! Electric field always points from higher electric potential to lower electric potential.

!! A positive charge accelerates from a region of higher electric potential energy (or higher potential) toward a region of lower electric potential energy (or lower potential).

!! A negative charge accelerates from a region of lower potential toward a region of higher potential.

Problem Solving Tactics

Example (1)

VAB =VB !VA =

"Uq

= !!E

A

B

# $d!l

VAB = !!E

A

B

" #d!l = !kQ dr

r2A

B

" = !kQ !1r

$%&

'()A

B

= kQ 1rB! 1rA

$%&

'()

Calculating Potential from the Field

dW =!F !d!l

!F = q0

!E

W = q0!E !d!l

i

f

"Potential difference is defined:

!V =Vf "Vi = "Wq0

Vf "Vi = "!E #d!l

i

f

$

Electric Potential Due to a Point Charge

Vab =Vb !Va = !!E "d!l

ra

rb#

Vb !Va = ! Q4$%0

1r2ra

rb# dr = 14$%0

Qrb! Qra

&'(

)*+

Setting the potential to zero at r = ! gives the general form of the potential due to a point charge:

Vb =14!"0

Qrb

Potential of a Collection of Point Charges

For a collection of point charges, the total electric potential at point P is sum of the potentials of each charge.

V =V1 +V2 +V3

V = k Q1r1+ k Q2

r2+ k Q3

r3

V = k Q1r1+ Q2r2

+ Q3r3

!"#

$%&

Example (2)

Show that the potential at point P in the figure is zero.

= ?

Potential due to Continuous Charge Distribution

Recall that the potential from a point charge is:

If we have a continuous charge distribution, then we break it up into small differential charges, dq, and sum them together by integrating over the volume.

V = k qr

dV = k dqr

V = k dqrfinite

!

782 C H A P T E R 2 5 Electric Potential

Px

!x2 + a2

dq

a

Figure 25.15 A uniformly charged ring of radius a lies in a planeperpendicular to the x axis. All segments dq of the ring are the samedistance from any point P lying on the x axis.

tion 25.16:

(25.21)

This result agrees with that obtained by direct integration(see Example 23.8). Note that at x ! 0 (the center ofthe ring). Could you have guessed this from Coulomb’s law?

Exercise What is the electric potential at the center of thering? What does the value of the field at the center tell youabout the value of V at the center?

Answer Because at the cen-Ex ! "dV/dx ! 0V ! keQ /a.

Ex ! 0

keQx(x2 # a2)3/2 !

! "keQ("12 )(x2 # a2)"3/2(2x)

Ex ! "dVdx

! "keQ ddx

(x2 # a2)"1/2

ter, V has either a maximum or minimum value; it is, in fact,a maximum.

Electric Potential Due to a Uniformly Charged DiskEXAMPLE 25.6from the definition of surface charge density (see Section23.5), we know that the charge on the ring is

Hence, the potential at the point P due tothis ring is

To find the total electric potential at P, we sum over all ringsmaking up the disk. That is, we integrate dV from r ! 0 to r ! a:

This integral is of the form un du and has the valuewhere and This gives

(25.22)

(b) As in Example 25.5, we can find the electric field atany axial point from

(25.23)

The calculation of V and E for an arbitrary point off the axisis more difficult to perform, and we do not treat this situationin this text.

2$ke % !1 "x

!x2 # a2"Ex ! "dVdx

!

2$ke %[(x2 # a2)1/2 " x]V !

u ! r 2 # x2.n ! "12un#1/(n # 1),

V ! $ke %#a

0

2r dr

!r 2 # x2! $ke %#a

0 (r 2 # x2)"1/2 2r dr

dV !ke dq

!r 2 # x2!

ke %2$r dr

!r 2 # x2

% dA ! %2$r dr.dq !

Find (a) the electric potential and (b) the magnitude of theelectric field along the perpendicular central axis of a uni-formly charged disk of radius a and surface charge density %.

Solution (a) Again, we choose the point P to be at a dis-tance x from the center of the disk and take the plane of thedisk to be perpendicular to the x axis. We can simplify theproblem by dividing the disk into a series of charged rings.The electric potential of each ring is given by Equation 25.20.Consider one such ring of radius r and width dr, as indicatedin Figure 25.16. The surface area of the ring is dA ! 2$r dr ;

Figure 25.16 A uniformly charged disk of radius a lies in a planeperpendicular to the x axis. The calculation of the electric potentialat any point P on the x axis is simplified by dividing the disk intomany rings each of area 2$r dr.

dr

dA = 2"rdr

! r 2 + x 2

x P

ra

"

Example (3) Find the electric potential at point P along the perpendicular central axis of a uniformly charged disk of radius a and surface charge density . !

Equipotential Surfaces

!V =Vf "Vi = "

!E #d!l

i

f

$

An equipotential is a line or surface over which the potential is constant.

Electric field lines are perpendicular to equipotential lines.

Point charge equipotential surfaces.

Equipotential Surfaces

The equipotential lines for the case of two equal and oppositely charged particles. Equipotential lines and surfaces are always perpendicular to field lines

The entire volume of a conductor must be at the same potential in the static case, so the surface of a conductor is an equipotential surface.

Electric Dipole Potential

The potential due to an electric dipole is just the sum of the potentials due to each charge, and can be calculated exactly. For distances large compared to the charge separation:

V = 14!"0

Qr+ 14!"0

#Q( )r + $r( ) =

14!"0

Q 1r# 1r + $r

%&'

()*

V = Q4!"0

$rr r + $r( )

r! l !"! #r $ l cos%r! #r = l cos%

V = 14&'0

Ql cos%r2 = 1

4&'0

pcos%r2

Calculating Field from the Potential

!V =Vf "Vi = "

!E #d!l

i

f

$

E = ! dVdl

The potential gradient gives the component of the field along the displacement .

Ex = ! "V"x

, Ey = ! "V"y

, Ez = ! "V"z

Example (4) An uncharged, infinitely long conducting cylinder of radius a is placed in an initially uniform electric field E=E0 i, such that the cylinder’s axis lies along the z-axis. The resulting electrostatic potential is:

Points inside the conductor : V x, y, z( ) =V0

Points outside the conductor : V x, y, z( ) =V0 ! E0x +E0a

2xx2 + y2

where, V0 is the electrostatic potential on the conductor. Determine the x, y, and z components of the resulting electric field.

Electrostatic Potential Energy: the Electron Volt

The joule is a very large unit for dealing with energies of electrons, atoms, or molecules. For this purpose, the unit electron volt (eV) is used. One electron volt is defined as the energy acquired by a particle carrying a charge whose magnitude equals that of the electron (q = e) as a result of moving through a potential difference of 1 V.

The charge on an electron has magnitude 1.6022 ! 10!19 C, and the change in potential energy equals qV. So

!U = qVAB Coulomb-Volts or Joules !

1 eV =(1.6022 ! 10!19 C)(1.00 V) = 1.6022 ! 10!19 J

Example (5)

What is the kinetic energy of a He2+ ion released from rest and accelerated through a potential difference of 2.5 kV?

a)! 2500 eV b)! 500 eV c)! 5000 eV d)! 10,000 eV e)! 250 eV

Electric Field & Potential

"! Inside conductor: E=0 and perpendicular to the surface "! Surface of any charged conductor in

equilibrium is an equipotential surface

Field Lines of Opposite Charges

Field Lines of Like Charges