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Potential Energy andthe Electric Potential

From last lecture:Positive and Negative chargesConductors, Insulators and SolutionsCharging processes (triboelectric, conduction,induction)Electric force (Coulomb’s law)

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Potential Energy andthe Electric Potential

This lecture:WorkConservative forces and potential energy

Electric Potential

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Recall from P207

For an infinitesimal displacement ds,the work done by the force F is:dW = F·ds = F ds cosθPositive: Force is in direction of motionNegative: Force is opposite to direction of motionZero: Force is perpendicular to direction of motion(eg gravitational force for a block moving on a plane)Along any path, a path integral is needed:

with K = kinetic energy

θ

F

ds

€

W = F ⋅dsA

B∫ = KB −KA

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Gravitational Force and work

WG=-mgh work done by the gravitational forceThe work done by an external force to lift the block of mass mis: W = -WG = Fh=mgh >0! Someone has to make an effort to lift the block!!

F

The work done by an external agent is equaland opposite to the work done by thegravitational force

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Conservative forces

Conservative Forces: the work done by the forceis independent on the path and depends only onthe starting and ending locations.

It is possible to define the potential energy U

Wconservative = −Δ U = Uinitial - Ufinal == -(Kfinal - Kinitial) = -ΔK

Fg

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Potential Energy of 2 charges

Consider 2 positive charged particles. The electric forcebetween them is

The work that an external agent should do to bring q2 at adistance rf from q1 starting from a very far away distance isequal and opposite to the work done by the electric force.Charges repel ⇒W>0!

r12

F

€

W = F ⋅dr∞

r12∫ = −WE

2

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Potential Energy of 2 charges

Since the 2 charges repel, the force on q2 due to q1

F12 is opposite to the direction of motion The external agent F = -F12 must do positive work!W > 0 and W=-WE

r12

Fdr

€

W = −keq1q2 −1r

∞

r12

= keq1q2r12€

W = −WE = − keq1q2r2∞

r12

∫ dr

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Potential Energy of 2 charges

The external agent changes the potential energy ofthe system W = ΔU = Ufinal - Uinitial

If W>0⇒U increaseas

We set Uinitial = U(∞) = 0 since at infinite distance theforce becomes null

The potential energy of the system is

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More About U of 2 Charges

Like charges ⇒ U > 0 and work must be doneto bring the charges together since they repel(W>0)

Unlike charges ⇒ U < 0 and work is done tokeep the charges apart since the attract one theother (W<0)

U and W (external agent against electricforce) have the same sign that is determinedby the product of the 2 charges!

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U with Multiple Charges

If there are more than twocharges, then find U foreach pair of charges andadd them

For three charges:

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Quick Quiz 1

−1µC

−3µC

−2µC5 m

5 m5 m

1. W = +19.8 mJ2. W = 0 mJ3. W = -19.8 mJ

Question: How much work would it take YOU toassemble 3 negative charges?

Likes repel, so YOU willstill do positive work!

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Work done to assemble 3 charges

W1 = 0

1µC

3µC

2µC5 m

5 m5 m

• W2 = k q1 q2 /r

• W3 = k q1 q3/r + k q2 q3/r = (9×109)(1×10-6)(3×10-6)/5 + (9×109)(2×10-6)(3×10-6)/5 =16.2 mJ

• W = +19.8 mJ• WE = -19.8 mJ• UE = +19.8 mJ

=(9×109)(1×10-6)(2×10-6)/5 =3.6 mJ

q3

q2q1

Similarly if they are all positive:

3

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Quick Quiz 2

+Q

+Q

− Q5 m

5 m5 m

1. positive2. zero3. negative

The total work required for YOU to assemble the set ofcharges as shown below is:

€

W1 = 0

W2 = k Q(−Q)d

W3 = k QQd

+ k Q(−Q)d

Total work = −k Q2

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Work and Δ Potential Energy

rf

• Charge moved ∞ → rf

• FE = kq1q2/r2

W = - =kq1q2/rf

ΔUE= +kq1q2/rf

€

k q1q2r2

dr∞

rf∫ =

Gravity Electric

• Mass moved ∞ → rf

• FG =Gm1m2/r2

• W=

= -Gm1m2/rf

ΔUG= -Gm1m2/rf

€

G m1m2

r2dr

∞

rf∫ =

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Electric Potential

The potential energy per unit charge, U/qo, is theelectric potential The potential is independent of the value of qo

The electric potential is

Since U is a scalar quantity also V is a scalar Just as with potential energy only differences are

meaningful It is customary to assume to choose a reference

potential of V = 0 at r = ∞16

Potential and Point Charges

The potential differencebetween points A and B is

Electric force conservative ⇒electric potential independent on pathbetween A and B

Units: 1 V = 1 J/1C

It takes 1 joule of work to move a 1C charge through apotential difference of 1 volt

€

VB −VA =UB

q0−UA

q0= keq

1rB−1rA

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Potential of Point Charges

Customary choise of reference potential:V = 0 at rA = ∞.

The electric potential due to several pointcharges is the sum of the potentials due to eachindividual charge (superposition principle)

€

VB −VA = keq1rB−1rA

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In the Figure, take q1 to be a negative source chargeand q2 to be a test charge. If q2 is initially positive andis changed to a charge of the same magnitude butnegative, the potential at the position of q2 due to q1

(a) increases

(b) decreases

(c) remains the same

Answer: (c). The potential is established only by the sourcecharge and is independent of the test charge.

Quick Quiz 3

_

+ → -

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Quick Quiz 4Consider the situation in QQ3 again: q1 is a negative source

charge and q2 a test charge. When q2 is changed frompositive to negative, the potential energy of the two-chargesystem

(a) increases

(b) decreases

(c) remains the same

Answer: (a). The potential energy of the two-charge system isinitially negative, due to the products of charges of opposite sign.When the sign of q2 is changed, both charges are negative, andthe potential energy of the system is positive.

_

+ → -

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Electrostatic machines to create potentialdifferences: Van de GraaffGenerator

The high-voltage electrode is a hollow metal domemounted on an insulated column

Charge is delivered continuously to dome by amoving belt of insulating material

Large potentials can be developed by repeated tripsof the belt

Protons can be accelerated through such largepotentials

€

ΔU = qΔV = −ΔK = K f

ΔV

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Toepler-Voss Electrostatic Machine(1860-80)

2 small metal brushes rub against the buttons; brushes are in contact withthe disks at opposite points

2 brass collection combs are at the extremes of the horizontal rod in frontof the buttons. They collect opposite charges

spherules of the spark-gapare charged by opposite signwith respect to the sign of thecomb with which they are incontact.The combs are in contact withthe inner shields of two Leidenjars

http://www.coe.ufrj.br/~acmq/electrostatic.html

ΔV

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Quick Quiz 5: Electric Dipole

€

V = keqiri

= keq

x − a−

qx + a

∑ =

2keqax 2 − a2

If P is far from the dipole x>>a

What if point P happens to be locatedto the left of -q?

€

V ≈2keqax 2

(*)

€

V = keqiri

= keq

x + a−

qx − a

∑ = −

2keqax 2 − a2

Answer: it has the same value but it’s negativep=qa

Electric dipole moment

For a point P on the right of +q on the x axis

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Verify that:

The general formula that applies to all casesalong the x axis is

What happens for x=0?€

V (x) = keq

x − a−

qx + a

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