Lecture 4 Work, Electric Potential and Potential Energy Ch. 25

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Lecture 4 Work, Electric Potential and Potential Energy Ch. 25

• Topics• Work, electric potential energy and electric potential• Calculation of potential from field• Potential from a point charge• Potential due to a group of point charges, electric dipole• Potential due to continuous charged distributions• Calculating the electric field from a potential• Electric potential energy from a system of point charges• Equipotential Surface• Potential of a charged isolated conductor

• Demos• Teflon and silk• Charge Tester, non-spherical conductor, compare charge density at different radii.

•Elmo•Potential in the center of four charges•Potential of a electric dipole

• Polling

2

W =

rF ⋅d

rs

i

f

∫ = qrE ⋅d

rs

i

f

∫

ΔU = −W

3

ΔV =ΔU

q

4

Work, Potential Energy and Electric potential• The electric force is mathematically the same as gravity so it too must be a conservative force.

We want to show that the work done is independent of the path and only depends on the endpoints. Then

the force is said to be a conservative force.

• First start with work Work done by the electric force =

• Then we will find it useful to define a potential energy as is the case for gravity.

• And the electric potential W =

rF ⋅d

rs

i

f

∫ = qrE ⋅d

rs

i

f

∫

ΔV =ΔU

q

ΔU = −W

5

Lets start with a uniform electric field and find the work done for a positive test charge.

b

a

E

P2 P3

P1

c

W = q

rE ⋅drs

i

f

∫ =qEcosθ s

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Find work done along path W12 for a positive test charge

b

a

E

P2 P3

P1

c

F

W12 =0W12 =qEcos90 a

W = q

rE ⋅drs

i

f

∫ =qEcosθ s

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Find Work along path W23

b

a

E

P2 P3

P1

c

F

W23 =qEb

W = q

rE ⋅drs

i

f

∫ =qEcosθ s

W23 =qEcos0 b

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W12 + W23 = 0 + qEb =qEb

Compare this work done along path W13

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Work done along path W13

b

a

P2 P3

P1

c

Fθ

W13 =qEcosθ c=qE

bc

c=qEb

W = q

rE ⋅drs

i

f

∫ =qEcosθ s

θ

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Conclusion• Work done along path W12 + W23 = W13.

• Work is independent of the particular path.• Although we proved it for a uniform field, it is true for any field that is a only a function of r and is along r.• It only depends on the end points i and f.

•This means we can define a function at every point in space and when we take the difference of that function between any two points, it is equal to the negative of the work done.

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b

a

P2 P3

P1

c

Fθ

When we go from P1 to P3 we evaluate the Work function at P3 and subtract the value at P1 and then the a difference equals the negative of the work done in going form P1 to P3. This function is called the potential energy function

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Example of finding the Potential Energy Function U in a Uniform Field

What is the electric potential difference for a unit positive charge moving in an uniform electric field from a to b?

E

Ed

a b

ΔU =Ub −Ua = −W = − q

rE ⋅d

rs

a

b

∫ = −Eq dxa

b

∫ = −Eq(xb − xa)

ab UUU −=Δ

If we set the origin at xb = 0, and measure from b to a, then

ΔU = qEd

This is analogy with gravitation where we U =mgh.

Ub=0 and the potential energy function is U=qEd

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Now define the Electric Potential Difference ΔV which does not depend on charge.

ΔV =ΔU

q

ΔU = −W

ΔV =−W

q0

• The potential difference is the negative of the work done per unit charge by an electric field on a positive unit charge when it moves from one point to another.

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ΔV = Ed

ΔV =ΔU

q

ΔV =qEd

q

For a battery of potential difference of 9 volts you

would say that the positive terminal is 9 volts above

the negative terminal.

ΔU = qEd

Find the potential difference ΔV for a uniform electric field

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Note relationship between potential and electric field

• V is a scalar not a vector. Simplifies solving problems.

• We are free to choose V to be 0 at any location. Normally V is chosen to be 0 at the negative terminal of a battery or 0 at infinity for a point charge.

ΔV =

−W

q0

= −

rF

q0

⋅drs∫ = −

rE ⋅d

rs∫ dV =−Edx

E =−dV / dx

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Example for a battery in a circuit• In a 9 volt battery, typically used in IC circuits, the

positive terminal has a potential 9 v higher than the negative terminal. If one micro-Coulomb of positive charge flows through an external circuit from the positive to negative terminal, how much has its potential energy been changed?

q

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Generalize concept of electric potential energy and potential

difference for any electric field

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(independent of path, ds)

Therefore, electric force is a conservative force.

= - Work done by the electric force = −

rF ⋅d

rs

i

f

∫ΔU =U f −Ui

qUV Δ=Δ

ΔV = Vf −Vi = −

rE ⋅d

rs∫

y

x

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Find the electric potential when moving from one point to another in a field due to a point charge?

ΔV = −

rE ⋅d

rr∫

rE =

kqr2 r

Vf −Vi =−

rE ⋅d

rr

i

f

∫

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Vf −Vi =−

rE ⋅d

rr

R

∞

∫ =−kqcos0o 1r2 d

R

∞

∫ r =kq1r R

∞

=kq(1∞−1R)

R

kqV =

04

1

πε=k

eqn 25-26

Replace R with r

V =1

4πε 0

qr

Vf −Vi =0−Vi =−kqR

Potential of a point charge at a distance R

Vf −Vi =−

rE ⋅d

rr

i

f

∫

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Electric potential for a positive point charge

V (r) =kqr

r = x2 + y2

• V is a scalar• V is positive for positive charges, negative for negative charges.

• r is always positive.

• For many point charges, the potential at a point in space is the simple algebraic sum (Not a vector sum)

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Hydrogen atom.

• What is the electric potential at a distance of 0.529 A from the proton? 1A= 10-10 m

Electric potential due to a positive point charge

r = 0.529 A

•What is the electric potential energy of the electron at that point?

V =kqR

=8.99 ×109 Nm2

C2( )×1.6 ×10−19C

.529 ×10−10m

V =27.2JC

=27.2Volts

U = qV= (-1.6 x 10-19 C) (27.2 V) = - 43.52 x 10-19 Jor - 27.2 eV where eV stands for electron volt.

Note that the total energy E of the electron in the ground state

of hydrogen is - 13.6 eV

Also U= 2E = -27.2 eV. This agrees with above formula.

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What is the electric potential due to several point charges?

For many point charges, the potential at a point in space is the simple algebraic sum (Not a vector sum)

⎟⎠

⎞⎜⎝

⎛ ++=3

3

2

2

1

1

r

q

r

q

r

qkV

V =kqi

rii∑

r1

xr3

y

r2

q1q2

q3

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Four Point charges

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What is the potential due to a dipole?

Two point charges that are opposite and equal

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Potential for a Continuous Distribution of Charge

Point charge

For an element of charge

Integrate

rkq

V =

rkdq

dVdq = ,

∫=r

kdqV

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Chaper 24 Problem 22. With V = 0 at infinity, what is the electric potential at P, the center of curvature of the uniformly charged nonconducting rod?

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Chapter 24 Problem 26. What is the magnitude of the net electric potential at the center? 1. A thin rod of charge -3.0 µC that forms a full circle of radius 6.0 cm2. A second thin rod of charge 2.0 µC that forms a circular arc of radius 4.0 cm, subtending an angle of 90° about the center of the full circle3. An electric dipole with a dipole moment that is perpendicular to a

radial line and that has magnitude 1.28 multiplied by 10-21 C·m

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Figure 24-44 shows a thin plastic rod of length L and uniform positive charge Q lying on an x axis. With V = 0 at infinity, find the electric potential at point P1 on the axis, at distance d from one end of the rod.

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Potential due to a ring of charge• Direct integration. Since V is a scalar, it is easier to evaluate V than E.

• Find V on the axis of a ring of total charge Q. Use the formula for a

point charge, but replace q with elemental charge dq and integrate.

Point charge

For an element of charge

r is a constant as we integrate.

This is simpler than finding E because V is not a vector.

rkq

V =

rkdq

dVdq = ,

∫=r

kdqV

Q)Rz(

kV22 +

=⇒∫+

= dqRz

k

)( 22

∫+

=)( 22 Rz

kdq

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Potential due to a line charge

rdq

kdV =We know that for an element of charge dq

the potential is

rdxkVd ,So λ=

For the line charge let the charge density be λ. Then dq=λdx

22 dxr ,But +=

22 dx

dxkVd ,Then+

λ=

Now, we can find the total potential V produced by the rod at point P by integrating along the length of the rod from x=0 to x=L

∫∫∫+

λ=+

λ==L

022

L

022

L

0 dx

dxkdx

dxk VdV ⇒ V = kλ ln(x + x2 + d 2 )0

L

)dln)dLL(ln(k Vo,S 22 −++λ= or, V =kλ lnL + L2 +d2

d

⎛

⎝⎜

⎞

⎠⎟

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A new method to find E if the potential is known. If we know V, how do we find E?

So the x component of E is the derivative of V with respect to x, etc.

–If V = a constant, then Ex = 0. The lines or surfaces on which V remains constant are called equipotential lines or surfaces.–See example on next slide

ΔV = −

rE ⋅d

rs∫

dV =−rE ⋅d

rs

dzdVE

dydVE

dxdVE

z

y

x

−=

−=

−=

dzEdyEdxEdV zyx −−−=

drs =idx+ jdy+ kdz

rE =Exi + Ey j + Ezk

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Now find the electric field at point P1 on the axis, at distance d from one end of the rod. Find the x and y components Ex and Ey.

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Equipotential Surfaces

• Three examples

• What is the equipotential surface and equipotential volume for an arbitrary shaped charged conductor?

• See physlet 9.3.2 Which equipotential surfaces fit the field lines?

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a) Uniform E field

V = constant in y and z directions

b) Point charge

(concentric shells)

c) Electric Dipole

(ellipsoidal concentric shells)

E =Ex,Ey =0,Ez =0

Ex =−dVdx

V =−Exd

Blue lines are the electric field lines

Orange dotted lines represent the equipotential surfaces

x

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Electric Potential Energy U of a system of charges

How much work is required to set up the arrangement of

Figure 24-46 if q = 3.20 pC, a = 54.0 cm, and the particles are

initially infinitely far apart and at rest?

q1q2

q3q4

W = U

U =k(q1q2

r12+

q1q3

r13+

q1q4

r14+

q2q3

r23+

q2q4

r24+

q3q4

r34)

U =k(−qqa

+qq2a

−qqa

−qqa

+qq2a

−qqa)

U =k(−4q2

a+2q2

2a)

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QuickTime™ and aTIFF (Uncompressed) decompressor

are needed to see this picture.

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Dielectric Breakdown: Application of Gauss’s Law

If the electric field in a gas exceeds a certain value, the gas breaks down and you get a spark or lightning bolt if the gas is air. In dry air at STP, you get a spark when

E ≥3×104 Vcm

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V = constant on surface of conductor Radius r2

r1

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This explains why:• Sharp points on conductors have the highest electric fields and cause

corona discharge or sparks.

• Pick up the most charge with charge tester from the pointy regions of the non-spherical conductor.

• Use non-spherical metal conductor charged with teflon rod. Show variation of charge across surface with charge tester.

Van de Graaff

+ + + +

- - - -

Radius R

1

2

V = constant on surface of conductor

Cloud

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How does a conductor shield the interior from an exterior electric field?

• Start out with a uniform electric field

with no excess charge on conductor.

Electrons on surface of conductor adjust

so that:

1. E=0 inside conductor

2. Electric field lines are perpendicular

to the surface. Suppose they weren’t?

3. Does E = just outside the conductor

4. Is uniform over the surface?

5. Is the surface an equipotential?

6. If the surface had an excess charge, how would your answers change?

0ε

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What is the electric potential of a uniformly charged circular disk?

We can treat the disk as a set of ring charges. The ring of radius R’ and thickness dR’ has an area of 2πR’dR’ and it’s charge is dq = dA = 2πR’)dR’ where =Q/(πR2), the surface charge density. The potential dV at point P due to the charge dq on this ring given by

Integrating R’ from R’=0 to R’=R

q

dV =kdq

(z2 + (R')2 )

dV =k2πR'dR'

(z2 _(R')2 )⇒

V =k2πR'dR'

(z2 + (R')2 )0

R

∫

⇒ V = 2kσπ ( z2 + R2 − z)

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Chapter 24 Problem 19

The ammonia molecule NH3 has a permanent electric dipole moment equal to 1.31 D, where 1 D = 1 debye unit = 3.34 multiplied by 10-30 C·m. Calculate the electric potential due to an ammonia molecule at a point 44.0 nm away along the axis of the dipole. (Set V = 0 at infinity.)

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Chapter 24 Problem 55

Two metal spheres, each of radius 1.0 cm, have a center-to-center separation of 2.2 m. Sphere 1 has charge +2.0 multiplied by 10-8 C. Sphere 2 has charge of -3.8 multiplied by 10-8 C. Assume that the separation is large enough for us to assume that the charge on each sphere is uniformly distributed (the spheres do not affect each other). Take V = 0 at infinity.

(a) Calculate the potential at the point halfway between the centers.(b) Calculate the potential on the surface of sphere 1.(c) Calculate the potential on the surface of sphere 2.

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Chapter 24 Problem 57

A metal sphere of radius 11 cm has a net charge of 2.0 multiplied by 10-8 C.

(a) What is the electric field at the sphere's surface?(b) If V = 0 at infinity, what is the electric potential at the sphere's surface?(c) At what distance from the sphere's surface has the electric potential decreased by 500 V?