Electric Potential II Physics 2102 Jonathan Dowling Physics 2102 Lecture 6.

Electric Potential IIElectric Potential II

Physics 2102

Jonathan Dowling

Physics 2102 Physics 2102 Lecture 6Lecture 6

ExampleExample

Positive and negative charges of equal magnitude Q are held in a circle of radius R.

1. What is the electric potential at the center of each circle?

• VA =

• VB =

• VC =

2. Draw an arrow representing the approximate direction of the electric field at the center of each circle.

3. Which system has the highest electric potential energy?

–Q

+Q

A

C

B€

k +3Q − 2Q( ) /r = +kQ /r

∑=i i

i

r

qkV

€

k +2Q − 4Q( ) /r = −2kQ /rk +2Q−2Q( ) / r =0

UB

Electric Potential of a Dipole (on axis)Electric Potential of a Dipole (on axis)

( ) ( )2 2

Q QV k k

a ar r

= −− +

What is V at a point at an axial distance r away from the midpoint of a dipole (on side of positive charge)?

a

r-Q +Q( ) ( )

2 2

( )( )2 2

a ar r

kQa a

r r

⎛ ⎞+ − −⎜ ⎟= ⎜ ⎟

⎜ ⎟− +⎝ ⎠

)4

(42

20

ar

Qa

−=

πε2

04 r

p

πε=

Far away, when r >> a:

V

Electric Potential on Perpendicular Electric Potential on Perpendicular Bisector of DipoleBisector of Dipole

You bring a charge of Qo = –3C from infinity to a point P on the perpendicular bisector of a dipole as shown. Is the work that you do:

a) Positive?

b) Negative?

c) Zero?

a

-Q +Q

-3C

P

U= QoV=Qo(–Q/d+Q/d)=0 

d

Continuous Charge Continuous Charge DistributionsDistributions

• Divide the charge distribution into differential elements

• Write down an expression for potential from a typical element — treat as point charge

• Integrate!• Simple example: circular

rod of radius r, total charge Q; find V at center.

dqV k

r=∫

dq

r

k Qdq k

r r= =∫

Potential of Continuous Charge Potential of Continuous Charge Distribution: Example Distribution: Example

/Q Lλ = dxdq λ=

∫∫ −+==

L

xaL

dxk

r

kdqV

0)(

λ

[ ]LxaLk 0)ln( −+−= λ

⎥⎦⎤

⎢⎣⎡ +=

a

aLkV lnλ

• Uniformly charged rod• Total charge Q• Length L• What is V at position P

shown?

Px

L a

dx

Electric Field & Potential: Electric Field & Potential: A Simple Relationship! A Simple Relationship!

Notice the following:

• Point charge:– E = kQ/r2

– V = kQ/r

• Dipole (far away):– E ~ kp/r3

– V ~ kp/r2

• E is given by a DERIVATIVE of V!

• Of course!

dx

dVEx −=

Focus only on a simple case:electric field that points along +x axis but whose magnitude varies with x.

Note: • MINUS sign!• Units for E -- VOLTS/METER (V/m)

f

i

V E dsΔ =− •∫r r

Electric Field & Potential: Example Electric Field & Potential: Example • Hollow metal sphere of

radius R has a charge +q• Which of the following is

the electric potential V as a function of distance r from center of sphere?

+q

Vr1≈

rr=R

(a)

Vr1≈

rr=R

(c)

Vr1≈

rr=R

(b)

+q

Outside the sphere:• Replace by point charge!Inside the sphere:• E =0 (Gauss’ Law) • E = –dV/dr = 0 IFF V=constant

2

dVE

drd Q

kdr r

Qk

r

=−

⎡ ⎤=− ⎢ ⎥⎣ ⎦

=V

r1≈

Electric Field & Potential: Example Electric Field & Potential: Example

E2

1r

≈

Equipotentials and ConductorsEquipotentials and Conductors• Conducting surfaces are

EQUIPOTENTIALs

• At surface of conductor, E is normal to surface

• Hence, no work needed to move a charge from one point on a conductor surface to another

• Equipotentials are normal to E, so they follow the shape of the conductor near the surface.

Conductors change the field Conductors change the field around them!around them!

An uncharged conductor:

A uniform electric field:

An uncharged conductor in the initially uniform electric field:

““Sharp”conductorsSharp”conductors• Charge density is higher at conductor

surfaces that have small radius of curvature

• E = ε0 for a conductor, hence STRONGER electric fields at sharply curved surfaces!

• Used for attracting or getting rid of charge: – lightning rods– Van de Graaf -- metal brush transfers

charge from rubber belt– Mars pathfinder mission -- tungsten points

used to get rid of accumulated charge on rover (electric breakdown on Mars occurs at ~100 V/m)

(NASA)

Summary:Summary:• Electric potential: work needed to bring +1C from infinity; units = V• Electric potential uniquely defined for every point in space --

independent of path!• Electric potential is a scalar -- add contributions from individual point

charges• We calculated the electric potential produced by a single charge:

V=kq/r, and by continuous charge distributions : V=∫ kdq/r• Electric field and electric potential: E= dV/dx• Electric potential energy: work used to build the system, charge by

charge. Use W=qV for each charge.• Conductors: the charges move to make their surface equipotentials. • Charge density and electric field are higher on sharp points of

conductors.