Lecture 7The Electric Potential

Electric Potential EnergyElectric Potential Energy

The electric force, like the The electric force, like the gravitational force, is a gravitational force, is a conservative force. conservative force. ((Conservative

force: The work is path-independent.))

As in mechanics, work isAs in mechanics, work is

Work done on the positive charge Work done on the positive charge by moving it from A to B by moving it from A to B

A B

E��������������

d cosW Fd

cosW Fd qEd

The work done by a conservative force equals the The work done by a conservative force equals the negativenegative of the change in potential energy, of the change in potential energy, PEPE

This equation is valid only for the case of a This equation is valid only for the case of a uniformuniform electric field electric field

PE W qEd

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

if d E

Question: In the figure, a proton moves Question: In the figure, a proton moves from point i to point f in a uniform from point i to point f in a uniform electric field directed as shown. Does electric field directed as shown. Does the electric field do positive, negative or the electric field do positive, negative or no work on the proton?no work on the proton?

A: positiveA: positiveB: negativeB: negativeC: no work is done on the protonC: no work is done on the protonSolution:Solution:B: negativeB: negative

The The potential differencepotential difference between points A and B, V between points A and B, VBB-V-VAA, , is defined as the change in potential energy (final minus is defined as the change in potential energy (final minus initial value) of a charge, q, moved from A to B, divided initial value) of a charge, q, moved from A to B, divided by the chargeby the charge

Electric potential is a Electric potential is a scalar quantityscalar quantityElectric potential difference is a measure of electric Electric potential difference is a measure of electric energy per unit chargeenergy per unit chargePotential is often referred to as “voltage”Potential is often referred to as “voltage”

B A

PEV V V

q

If the work done by the electric field is zero, then the electric potential must be constant

Electric potential difference is the work done to move a charge from Electric potential difference is the work done to move a charge from a point A to a point B divided by the magnitude of the charge. Thus a point A to a point B divided by the magnitude of the charge. Thus the SI units of electric potential differencethe SI units of electric potential difference

In other words, 1 J of work is required to move a 1 C of charge In other words, 1 J of work is required to move a 1 C of charge between two points that are at potential difference of 1 Vbetween two points that are at potential difference of 1 V

Question: How can a bird stand on a high voltage line without Question: How can a bird stand on a high voltage line without getting zapped? getting zapped?

1 1V J C

Units of electric field (N/C) can be expressed in terms of the Units of electric field (N/C) can be expressed in terms of the units of potential (as volts per meter)units of potential (as volts per meter)

Because the positive tends to move in the direction of the Because the positive tends to move in the direction of the electric field, work must be done on the charge to move it in electric field, work must be done on the charge to move it in the direction, opposite the field. Thus,the direction, opposite the field. Thus,

A positive charge gains electric potential energy when it is moved in A positive charge gains electric potential energy when it is moved in a direction opposite the electric fielda direction opposite the electric field

A negative charge looses electrical potential energy when it moves A negative charge looses electrical potential energy when it moves in the direction opposite the electric fieldin the direction opposite the electric field

1 1N C V m

Example : Example : A uniform electric field of magnitude 250 V/m is directed in the A uniform electric field of magnitude 250 V/m is directed in the positive x direction. A +12µC charge moves from the origin to the point (x,y) positive x direction. A +12µC charge moves from the origin to the point (x,y) = (20cm, 50cm). (a) What was the change in the potential energy of this = (20cm, 50cm). (a) What was the change in the potential energy of this charge? (b) Through what potential difference did the charge move? charge? (b) Through what potential difference did the charge move?

Begin by drawing a picture of the situation, including the direction of the Begin by drawing a picture of the situation, including the direction of the electric field, and the start and end point of the motion. electric field, and the start and end point of the motion.

(a) The change potential energy is given by the charge times the field times (a) The change potential energy is given by the charge times the field times the distance moved parallel to the field. Although the charge moves 50cm the distance moved parallel to the field. Although the charge moves 50cm in the y direction, the y direction is perpendicular to the field. Only the 20cm in the y direction, the y direction is perpendicular to the field. Only the 20cm moved parallel to the field in the x direction matters for determining the moved parallel to the field in the x direction matters for determining the change of potential energy. PE = -qEd = -(+12µC)(250 V/m)(0.20 m) change of potential energy. PE = -qEd = -(+12µC)(250 V/m)(0.20 m) = -6.0×10 = -6.0×10-4-4 . .

(b) The potential difference iS the difference of electric potential,. (b) The potential difference iS the difference of electric potential,. V = PE / q = -6.0×10-4 J / 12µC = -50 V. V = PE / q = -6.0×10-4 J / 12µC = -50 V.

Analogy between electric and gravitational fieldsAnalogy between electric and gravitational fields

The same kinetic-potential energy theorem works hereThe same kinetic-potential energy theorem works here

If a positive charge is released from A, it accelerates in If a positive charge is released from A, it accelerates in the direction of electric field, i.e. gains kinetic energythe direction of electric field, i.e. gains kinetic energyIf a negative charge is released from A, it accelerates in If a negative charge is released from A, it accelerates in the direction opposite the electric fieldthe direction opposite the electric field

A

B

qd

A

B

mdE

��������������g��������������

i i f fKE PE KE PE

Example: motion of an electronExample: motion of an electron

Vab

What is the speed of an electron accelerated from rest across a potential difference of 100V?

Given:

V=100 Vme = 9.1110-31 kg

mp = 1.6710-27 kg

|e| = 1.6010-19 C

Find:ve=?vp=?

i i f fKE PE KE PE

f i fKE KE KE PE q V

21 2

2 f f

q Vmv q V v

m

6 55.9 10 , 1.3 10e pm mv vs s

Problem: Problem: A proton is placed between two parallel conducting plates in a A proton is placed between two parallel conducting plates in a

vacuum as shown.vacuum as shown. The potential difference between the two plates is 450 V. The The potential difference between the two plates is 450 V. The

proton is released from rest close to the positive plate.proton is released from rest close to the positive plate. What is the kinetic energy of the proton when it reaches the What is the kinetic energy of the proton when it reaches the

negative plate?negative plate?

+ -

Example (a) Through what potential difference would an electron need to accelerate to achieve a speed of 60% of the speed of light, starting from rest? (The speed of light is 3.00×108 m/s.)

(a) The final speed of the electron is vf = 0.6(3.00×108 m/s) = 1.80×108 m/s. At this speed, the energy (non-relativistic) is the kinetic energy,E = ½mvf² = 0.5(9.11×10-31 kg)(1.80×108 m/s)² = 1.48×10-14 J.This energy must equal the change in potential energy from moving through a potential difference, V, E = PE = qV. Therefore:V = E/q = (1.48×10-14 J)/(-1.6×10-19 C) = -9.25×104 V.

Electric potential and potential energy due to Electric potential and potential energy due to point chargespoint charges

Electric circuits: point of zero potential is defined by Electric circuits: point of zero potential is defined by grounding some point in the circuitgrounding some point in the circuitElectric potential due to a point charge at a point in Electric potential due to a point charge at a point in space: point of zero potential is taken at an infinite space: point of zero potential is taken at an infinite distance from the chargedistance from the chargeWith this choice, a potential can be found asWith this choice, a potential can be found as

Note: the potential depends only on charge of an object, Note: the potential depends only on charge of an object, qq, and a distance from this object to a point in space, , and a distance from this object to a point in space, rr..

e

qV k

r

Superposition principle for potentialsSuperposition principle for potentials

If more than one point charge is present, their electric If more than one point charge is present, their electric potential can be found by applying potential can be found by applying superposition superposition principleprinciple

The total electric potential at some point P due to several The total electric potential at some point P due to several point charges is the algebraic sum of the electric point charges is the algebraic sum of the electric potentials due to the individual charges.potentials due to the individual charges.

Remember that potentials are scalar quantities!Remember that potentials are scalar quantities!

Potential energy of a system of point Potential energy of a system of point chargescharges

Consider a system of two particlesConsider a system of two particles

If VIf V11 is the electric potential due to charge q is the electric potential due to charge q11 at a point P, at a point P,

then work required to bring the charge qthen work required to bring the charge q22 from infinity to P from infinity to P

without acceleration is qwithout acceleration is q22VV11. If a distance between P and . If a distance between P and

qq11 is r, then by definition is r, then by definition

Potential energy is Potential energy is positivepositive if charges are of the if charges are of the same same signsign and vice versa. and vice versa.

P A

q1q2

r

1 22 1 e

q qPE q V k

r

Example: potential energy of an ionExample: potential energy of an ion

Three ions, Na+, Na+, and Cl-, located such, that they form corners of an equilateral triangle of side 2 nm in water. What is the electric potential energy of one of the Na+ ions?

Cl-

Na+ Na+

? Na Cl Na Na Na

e e e Cl Na

q q q q qPE k k k q q

r r r

but : !Cl Naq q

0Nae Na Na

qPE k q q

r

Potentials and charged conductorsPotentials and charged conductors

Recall that work is opposite of the change in potential Recall that work is opposite of the change in potential energy,energy,

No work is required to move a charge between two points No work is required to move a charge between two points that are at the same potential. That is, W=0 if Vthat are at the same potential. That is, W=0 if VBB=V=VA A

Recall: Recall: 1.1. all charge of the charged conductor is located on its surfaceall charge of the charged conductor is located on its surface

2.2. electric field, E, is always perpendicular to its surface, i.e. no work is electric field, E, is always perpendicular to its surface, i.e. no work is done if charges are moved along the surfacedone if charges are moved along the surface

Thus: potential is constant everywhere on the surface of a Thus: potential is constant everywhere on the surface of a charged conductor in equilibriumcharged conductor in equilibrium

B AW PE q V V

… but that’s not all!

Because the electric field in zero inside the conductor, Because the electric field in zero inside the conductor, no work is required to move charges between any two no work is required to move charges between any two points, i.e. points, i.e.

If work is zero, any two points inside the conductor have If work is zero, any two points inside the conductor have the same potential, i.e. potential is constant everywhere the same potential, i.e. potential is constant everywhere inside a conductorinside a conductor

Finally, since one of the points can be arbitrarily close to Finally, since one of the points can be arbitrarily close to the surface of the conductor, the surface of the conductor, the electric potential is the electric potential is constant everywhere inside a conductor and equal to its constant everywhere inside a conductor and equal to its value at the surfacevalue at the surface!!

Note that the potential inside a conductor is Note that the potential inside a conductor is notnot necessarily zero, necessarily zero, even though the interior electric field is always zero!even though the interior electric field is always zero!

0B AW q V V

The electron voltThe electron volt

A unit of energy commonly used in atomic, nuclear and A unit of energy commonly used in atomic, nuclear and particle physics is electron volt (eV)particle physics is electron volt (eV)

The electron volt is defined as the energy that electron The electron volt is defined as the energy that electron (or proton) gains when accelerating through a potential (or proton) gains when accelerating through a potential difference of 1 Vdifference of 1 V

Relation to SI:Relation to SI:

1 eV = 1.601 eV = 1.601010-19 -19 CC··V = 1.60V = 1.601010-19 -19 J J

Vab=1 V

Example : ionization energy of the electron Example : ionization energy of the electron in a hydrogen atomin a hydrogen atom

In the Bohr model of a hydrogen atom, the electron, if it is in the ground state, orbits the proton at a distance of r = 5.2910-11 m. Find the ionization energy of the atom, i.e. the energy required to remove the electron from the atom.

Note that the Bohr model, the idea of electrons as tiny balls orbiting the nucleus, is not a very good model of the atom. A better picture is one in which the electron is spread out around the nucleus in a cloud of varying density; however, the Bohr model does give the right answer for the ionization energy

In the Bohr model of a hydrogen atom, the electron, if it is in the ground state, orbits the proton at a distance of r = 5.29 x 10-11 m. Find the ionization energy, i.e. the energy required to remove the electron from the atom.

Given:

r = 5.292 x 10-11 mme = 9.1110-31 kg

mp = 1.6710-27 kg

|e| = 1.6010-19 C

Find:

E=?

The ionization energy equals to the total energy of the electron-proton system,

E PE KE

22 2182.18 10 J -13.6 eV

2 2e

e e

k ee m eE k k

r mr r

The velocity of e can be found by analyzing the force on the electron. This force is the Coulomb force; because the electron travels in a circular orbit, the acceleration will be the centripetal acceleration:

c cma F

2 2

,2e

e vPE k KE m

r with

or2 2

2,e

v em k

r r or

22 ,e

ev k

mr

Thus, total energy is

Calculating the Potential from the FieldCalculating the Potential from the Field

To calculate the electric potential from the electric field we start with To calculate the electric potential from the electric field we start with the definition of the the definition of the work work dWdW done on a particle with charge done on a particle with charge qq by a by a force force FF over a over a displacement displacement dsds

In this case the force is provided by the electric fieldIn this case the force is provided by the electric fieldFF = = qEqE

Integrating the work done by the electric force on the particle as it Integrating the work done by the electric force on the particle as it moves in the electric field from some initial point moves in the electric field from some initial point ii to some final point to some final point ff we obtain we obtain

sdFdW

sdEqdW

f

isdEqW

Calculating the Potential from the FieldCalculating the Potential from the Field

Remembering the relation between the change in electric potential Remembering the relation between the change in electric potential and the work done …and the work done …

……we findwe find

Taking the convention that the electric potential is zero at infinity we Taking the convention that the electric potential is zero at infinity we can express the electric potential in terms of the electric field ascan express the electric potential in terms of the electric field as

qW

V e

f

i

eif sdE

qW

VVV

i

sdExV

)( ( i = , f = x)

Example - Charge moves in E fieldExample - Charge moves in E field

Given the uniform electric field E, find the Given the uniform electric field E, find the potential difference Vpotential difference Vff-V-Vi i by moving a test by moving a test

charge qcharge q00 along the path icf. along the path icf.

Idea: Idea: Integrate Integrate EEdsds along the path along the path connecting ic then cf. (Imagine that we connecting ic then cf. (Imagine that we move a test charge qmove a test charge q00 from i to c and then from i to c and then

from c to f.) from c to f.)

Example - Charge moves in E fieldExample - Charge moves in E field

EdVV

EdsEsdE

sdE

sdEsdEVV

if

f

c

f

c

c

i

f

c

c

iif

21distance)45cos(

E)lar toperpendicu (ds 0

distance = sqrt(2) d by Pythagoras

QuestionQuestion

We just derived VWe just derived Vff-V-Vii for the path for the path i -> c -> fi -> c -> f. What is V. What is Vff--

VVii when going directly from when going directly from ii to to ff ? ?

A: 0A: 0

B: -EdB: -Ed

C: +EdC: +Ed

D: -1/2 EdD: -1/2 Ed

Quick: V is independent of path.Explicit: V = - E . ds = E ds = - Ed

Example 1: What potential difference is needed to stop an electron with an initial speed of 4.2*105m/s?

W = Δk = kf - ki

W = 0-ki

q V = -1/2 mv2

V = -(1/2 mv2) / qV = -(1/2 x 9.1x10-31x(4.2x105)2)/1.5x10-19

V = -0.57 volt

Example 2: An ion accelerated through a potential difference of 115V experiences an increase in potential energy of 7.37*10-17J. Calculate the

charge on the ion.

VB - VA = Wab/ qq = Wab/ VB - VA

q= 6.41x10-19C

Examples & Problems

Example 3: An infinite charged sheet has a surface charge density s of 1.0*10-7 C/m2.  How far apart are the equipotential surfaces whose

potentials differ by 5.0 V?

                  E = 1.010-7/(2x8.85x10-12)= 5650

N/CV = Edd = V/Ed = 8.8x10-4m

Example 4: At what distance from a point charge of 8mC would the potential equal 3.6*104V?

  r = 2 m

Example 5: At a distance r away from a point charge q, the electrical potential is V=400v and the magnitude of the electric field is E=150N/C. 

Determine the value of q and r.

r = 2.67m

Example 6: Calculate the value of the electric potential at point P due to the charge configuration shown in Figure 5.19. Use the values q1=5C,

q2=-10C, a=0.4m, and b=0.5m.                                                                 

V = V1 + V2 + V3 + V4

Example 7: Two large parallel conducting plates are 10cm apart and carry equal and opposite charges on their facing surfaces.  An electron placed midway between the two plates experiences a force of 1.6*1015N.  What is

the potential difference between the plates?

V=EdE = F/qV = (F/q) dV = (1.6x1015/1.6x10-19) x 0.05 = 5x1032v

Example 8: Two point charges are located as shown in, where ql=+4C,

q2=-2C, a=0.30m, and b=0.90m. Calculate the value of the electrical

potential at points P1, and P2.  Which point is at the higher potential?                                                                   

Vp1 = V1 + V2

Example 9:  In figure 5.22 prove that the work required to put four charges together on the corner of a square of radius a is given by  (w=-0.21q2/oa).

U = U12 + U13 + U14 + U23 + U24 + U34

                                                                                                                                                          

ExampleExample

Assume we have a system of three Assume we have a system of three point charges:point charges:qq11 = +1.50 = +1.50 CC

qq22 = +2.50 = +2.50 CC

qq33 = -3.50 = -3.50 C.C.

qq11 is located at (0,a) is located at (0,a)

qq22 is located at (0,0) is located at (0,0)

qq33 is located at ( is located at (bb,0),0)

aa = 8.00 m and = 8.00 m and bb = 6.00 m. = 6.00 m.

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

The electric potential at point P is given The electric potential at point P is given by the sum of the electric potential from by the sum of the electric potential from the three chargesthe three charges

V kqi

rii1

3

kq1

r1

q2

r2

q3

r3

k

q1

b

q2

a2 b2

q3

a

V 8.99 109 N/C 1.50 10 6 C

6.00 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

r1

r2

r3

(1) Two charges q=+2*10-6C are fixed in space a distance d=2cm apart, as shown in figure (a) What is the electric potential at point C? (b) You bring a

third charge qn=2.0*10-6C very slowly from infinity to C.  How much work must you do?  (c) What is the potential energy U of the configuration when

the third charge is in place?

(2) Four equal point charges of charge q=+5mC are located at the corners of a 30cm by 40cm rectangle. Calculate the electric potential energy stored in

this charge configuration.

(3) Two point charges, Q1=+5nC and Q2=-3nC, are separated by 35cm. (a) What is the potential energy of the pair?  What is the significance of the

algebraic sign of your answer? (b) What is the electric potential at a point midway between the charges?

(4) What is the potential at the center of the square shown in figure 5.9? Assume that  q1= +1 ´10-8C,  q2= -2´10-8C,  q3=+3´10-8C, q4=+2´10-8C, and a=1m.

(5) Two charges of 2C and -6C are located at positions (0,0) m and (0,3) m, respectively as shown in figure.  (i) Find the total electric potential due to these charges at point (4,0) m.  (ii) How much work is required to bring a 3C charge from to the point P?  (iii) What is the potential energy for the three charges?

(6) A particle having a charge q=3*10-9C moves from point a to point b along a straight line, a total distance d=0.5m.  The electric field is uniform

along this line, in the direction from a to b, with magnitude E=200N/C.  Determine the force on q, the work done on it by the electric field, and the

potential difference Va-Vb.

(7) For the charge configuration shown in figure , Show that V(r) for the points on the vertical axis, assuming r >> a, is given by