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PHY2049: Chapter 24 22 ConcepTest: Electric Potential Which point has the largest potential when Q > 0? Which two points have the same potential? (a) A and C (b) B and E (c) B and D (d) C and D (e) no pair A C B D E Q E Smallest radius Same radius

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PHY2049: Chapter 24 22

ConcepTest: Electric Potential Which point has the largest potential when Q > 0?

Which two points have the same potential?(a) A and C(b) B and E(c) B and D(d) C and D(e) no pair

AC

BD

E

Q

E

Smallest radius

Same radius

PHY2049: Chapter 24 23

Multiple Charges: Superposition3 charges: Find total potential at a point in space

− Q3

+Q1

+Q2

xr1

r3r2

31 21 2 3

1 2 3tot

QQ QV V V V k k kr r r

−= + + = + +

V is a scalarNo directions to worry about!But you do have to watch signs!

+Q1

+Q2

−Q3

PHY2049: Chapter 24 24

ConcepTest: Electric Potential What is V at point A?

(a) V > 0 (b) V = 0 (c) V < 0

What is V at point B?(a) V > 0(b) V = 0(c) V < 0

Closer to + charge

Equal distance to both charges

A B

Q2 = +50μC Q1 = −50μC

30 cm

26 cm 26 cm

40 cm

60 cm

PHY2049: Chapter 24 25

ConcepTest: Electric Potential Which configuration gives V = 0 at all points on x axis?

(a) A(b) B(c) C(d) All of the above (e) None of the above

A

x

+2μC

-2μC

+1μC

-1μCB

x

+2μC

-1μC

+1μC

-2μCC

x

+2μC

-1μC

-2μC

+1μC

All points on x axis equidistantfrom each pair of charges

PHY2049: Chapter 24 26

ConcepTest: Electric Potential At which point does V = 0?

(a) C (b) A (c) D (d) B (e) all of the above

A

C

B

D

+Q −QAll points equidistantfrom charges

PHY2049: Chapter 24 27

ConcepTest: Fields & Potentials Find E and V at the center of the square.

(a) E = 0 V = 0(b) E = 0 V ≠ 0(c) E ≠ 0 V ≠ 0(d) E ≠ 0 V = 0(e) E = V regardless of the value

-Q

-Q +Q

+Q

PHY2049: Chapter 24 28

ConcepTest: Electric Potential You move a positive charge Q from A to B along the path shown. What is the sign of the work done by you?

(a) WAB < 0(b) WAB = 0(c) WAB > 0

A B

No change in potential sincedistance from center is the same

PHY2049: Chapter 24 29

Calculating E From Electric Potential VChange in potential from small change in distance

,y

x z

VEy

∂= −

x y zdV d E dx E dy E dz= − ⋅ = − − −E s

,x

y z

VEx

∂= −

,z

x y

VEz

∂= −

“Partial derivative wrt x” ⇒ y,z const

“Partial derivative wrt y” ⇒ x,z const

“Partial derivative wrt z” ⇒ x,y const

PHY2049: Chapter 24 30

Example: V = 30 – 5x

Equipotential Surfaces:Planes @ constant x

V = 30 25 20 15 10

x = 0 1 2 3 4

30 5V x= −

PHY2049: Chapter 24 31

Find E Field for V = 30 – 5xV = 30 25 20 15 10

x = 0 1 2 3 4

Differentiate V to get Ex, Ey, Ez

30 5

5

0

0

x

y

z

V xVExVEyVEz

= −∂

= − =∂∂

= − =∂∂

= − =∂

So E = (5,0,0)

PHY2049: Chapter 24 32

Example: Electric Field of Point ChargePoint charge

2 2ˆ, ,kq x y z kqE r

r r rr r⎛ ⎞= ≡⎜ ⎟⎝ ⎠

( ) ( )1/ 2 3/ 2 32 2 2 2 2 2

3 3likewise

x

y z

kq kqx kqxEx rx y z x y z

kqy kqzE Er r

∂= − = =

∂ + + + +

= =

2 2 2r x y z= + +

Coulomb’s law, as expected

kqVr

=

PHY2049: Chapter 24 33

Potential of Charge DistributionGeneralize superposition to continuous distribution

Distribution can be any shapeLine, surface, volume

Express dq in terms of charge densityLine or arc: dq = λds or dq = λdx (λ = linear charge density)Surface: dq = σdA (σ = surface charge density)Volume: dq = ρdV (ρ = volume charge density)

Express r in terms of a problem’s “natural” coordinatesx, θ, r, …

totkdqVr

= ∫

PHY2049: Chapter 24 34

Example: Charged RingFind V at a point z above axis of charged ring of radius R

( )2

0 2 2

k RdkdqVr z R

π λ θ= =

+∫ ∫

2Q

π=

2 2 2 2

2 k R kQVz R z R

π λ= =

+ +

z

( )3/ 22 2z

V kQzEz z R

∂= − =

∂ +

2 2

dq Rd

r z R

λ θ=

= +

R

r

2For zkQz R Ez

=

PHY2049: Chapter 24 35

Example: Charged LineFind V above midpoint of line of charge Q, length L

( )/ 2

/ 2 2 2

L

L

k dxkdqVr y x

λ−

= =+

∫ ∫

/Q Lλ =

2 2

2 2

/ 4 / 2ln

/ 4 / 2

y L Lk

y L Lλ

⎛ ⎞+ +⎜ ⎟=⎜ ⎟+ −⎝ ⎠

y

P

xL dq dxλ=

2 2r x y= +

PHY2049: Chapter 24 36

Charged Line: Limit of L y

Rationalize expression inside ln()

For L y2

2ln Ly

⎛ ⎞→ ⎜ ⎟⎜ ⎟

⎝ ⎠

( )22 22 2

22 2

/ 4 / 2/ 4 / 2ln ln

/ 4 / 2

y L Ly L Lyy L L

⎡ ⎤+ +⎛ ⎞ ⎢ ⎥+ +⎜ ⎟ = ⎢ ⎥

⎜ ⎟ ⎢ ⎥+ −⎝ ⎠ ⎢ ⎥⎣ ⎦

2 ln LV ky

λ⎛ ⎞

= ⎜ ⎟⎝ ⎠

PHY2049: Chapter 24 37

Charged Line (cont)Calculate y component of electric field at midpoint

2 2

2 2

/ 4 / 2ln

/ 4 / 2

y L LV k

y L Lλ

⎛ ⎞+ +⎜ ⎟=⎜ ⎟+ −⎝ ⎠

2 2 / 4y

V k LEy y y L

λ∂= − =

∂ +Agrees with calculationin chapter 22

PHY2049: Chapter 24 38

Example: Charged DiskFind V at a point z above axis of charged disk of radius R

( )0 2 2

2R k dkdqVr z

σ πρ ρ

ρ= =

+∫ ∫

( )2 22V k z R zπ σ= + −

2 (surface charge density)QR

σπ

=

( )2dq dA dσ σ πρ ρ= =

z

R

r

ρ

PHY2049: Chapter 24 39

Charged Disk (cont)Calculate z component of electric field

When z very small

( )2 22V k z R zπ σ= + −

2 22 1z

V zE kz z R

π σ⎛ ⎞∂

= − = −⎜ ⎟⎜ ⎟∂ +⎝ ⎠

02

2zE k σπ σε

= Just like sheet of chargeusing Gauss’ law

PHY2049: Chapter 24 40

Dipole

Like charges

Where are equilibrium points?

V

x

−Q

+Q

V

x+Q+Q

1 2

kQ kQVr r

= − +

1 2

kQ kQVr r

= +

No equilibrium since E is never 0

Equilibrium is at x = 0, since E = 0E is – dV/dx

PHY2049: Chapter 24 41

Conductors are EquipotentialsNo work to move along conductor

W = 0 = −qΔVAB ⇒ V is constant in conductor

But E = 0 inside surface bounded by conductorVC = VA ⇒ V is constant within enclosed volume

A

C

B

PHY2049: Chapter 24 42

Conductors in Electrostatic EquilibriumElectric field is zero everywhere inside the conductor

if E ≠ 0, then charges would move – no equilibrium!!

Excess charge on isolated conductor is only on surfaceMutual repulsion pushes the charges apart

Electric field is perpendicular to the surface of a conductorIf a parallel component existed, charges would move!!

For irregular shaped conductors, charge density is highest near sharp points, i.e. the field strength is greater there

PHY2049: Chapter 24 43

Spherical Shell

+Q

-

-

--

-

--

-- -

++

+

++

++

+

+

+ What is charge on inner shell?−Q

What is charge on outer shell?+Q

What is V vs radius?Constant from 0 < r < r2Falls as kQ / r for r > r2Inner radius = r1

Outer radius = r2

PHY2049: Chapter 24 44

A positively charged rod is held near a neutral conducting sphere. A positively charged particle is moved from A to B (A, B both on sphere).

The mechanical work required to cause this motion is (a) positive(b) zero(c) negative(d) depends on the path taken from A to B(e) cannot be determined without more information

ConcepTest: Electric Energy

All points on sphere are at same potential

PHY2049: Chapter 24 45

ConcepTest: Electric EnergyA positively charged rod is held near a neutral conducting sphere. A positively charged particle is moved from A to B (A is on sphere).

The mechanical work required to cause this motion is (a) positive(b) zero(c) negative(d) depends on the path taken from A to B(e) cannot be determined without more information

Must push against electrostatic force

PHY2049: Chapter 24 46

A positively charged rod is held near a neutral conducting sphere. A positively charged particle is moved from A to B(A on sphere).

The electrostatic work done on the particle is(a) positive(b) zero(c) negative(d) depends on the path taken from A to B(e) cannot be determined without more information

ConcepTest: Electric Energy

Electrostatic force is against direction of motion

PHY2049: Chapter 24 47

A positively charged rod is held near a neutral conducting sphere (A on sphere).

The potential change from A to B is:(a) positive(b) zero(c) negative(d) depends on the path taken from A to B(e) cannot be determined without more information

ConcepTest: Electric Potential

Higher potential near + charge

PHY2049: Chapter 24 48

Two charged metal spheres are connected by a copper wire. Note that rA > rB.

Which quantity must be the same for both spheres?(a) potential at the surface(b) charge on the sphere(c) surface charge density(d) field at the surface(e) more than one of the above.

ConcepTest: Electrostatics

PHY2049: Chapter 24 49

Two charged metal spheres are connected by a copper wire. Note that rA > rB.

Compare qA to qB(a) qA > qB

(b) qA < qB

(c) qA = qB

(d) Need more information

ConcepTest: Electrostatics

Potential is same, so kqA/rA = kqB/rB

PHY2049: Chapter 24 50

A solid spherical conductor is given a net nonzero charge. The electric potential of the conductor is

(a) largest at the center. (b) largest on the surface. (c) largest somewhere between center and surface. (d) constant throughout the volume.

ConcepTest: Electric Potential

PHY2049: Chapter 24 51

Review: Electric Potential

What charge will make the potential zero at X ?

What charge will make thepotential zero at X?

What charge will make thepotential zero at X?

-2Q +5Q

Charge = ??

x

+4Q

-4Q +Q

Charge = ??

x

+3Q-Q

r r

2r

Charge = ??

x

PHY2049: Chapter 24 52