1. ElectrostaticsElectric Potential

Recall…

Gravitational Potential Energy

or

Elastic Potential Energy

Now…

-+

+++++

+

++

+ +

Electric Potential Energy

(EPE)

EPE is a type of mechanical energy, like…

Kinetic Energy (KE) = ½ mv2

Rotational Kinetic Energy (KER) = ½ I2

Gravitational Potential Energy (PEgrav) = mgh

Elastic Potential Energy(PEelast) = ½ kx2

= Total Mechanical Energy (E)is conserved if there are no non-conservative forces present (ie friction).

+++

+

F = mg

q

F = qE

W = mgh W = qEx

∆Ug = -W ∆Ue = -W

b

m

h g

a

E x

Gravitational field Electric field

W= - (Fcos)s

Two points are said to differ in electric potential if work is done to move a charge from one point to another point in

an electric field. Work = -Δ Electric Potential Energy

Work is required to move two point charge closer together.

F Fr

This work is converted to potential energy.

This electric potential energy of two point charges: r

QQkU 21

e

[3] An electron starts from rest 32.5 cm from a fixed point charge with Q = -0.125 μC How fast will the electron be moving when it is very far away?

finalinitial KEPE

2mv

r

QQk

221

mr

QkQ2v 21

m 0.325gk 10x 9.1

C 10x 1.25 C 10x 1.6/CmN 10x 8.992v

13

719229

/sm 10x 49.3v 7

oo q

EPEVor

q

WV Is the potential energy per

unit charge.

E

q

x

a

b

q

UV a

a

q

UV b

b

o

AB

o

A

o

BABBA q

W

q

EPE

q

EPEVVV

o

AB

o q

W

q

EPEV

Electrical Potential can also be described by the terms; potential difference, voltage, potential drop, potential rise, electromotive force, and EMF

Units ~ J/C

Units for Electric Potential

• Electric potential V has units of Joules/Coulomb which is defined as a Volt:

1 Volt = 1 Joule/Coulomb• One Joule is the work done in moving one Coulomb

of charge through a potential difference of one Volt.• Electric field has units of Newtons/Coulomb or

Volts/meter:1N/C = 1 J/(m C) = 1 V/m

ABAB VVqUUU Difference inelectric potential

Difference inpotential energy

Potential V is the analog of height/level/altitude/elevation h.

If a charged particle gains kinetic energy in an electric field, it loses an equivalent amount of potential energy.

Point B is at a lower electric potential than is point A.

Points B and C are at the same electric potential.

The gain in kinetic energy depends only on the potential difference and not the path taken.

Example: Kinetic energy of an electron accelerated through a potential difference of 6,000 Volts (6 kV):

1 electron-volt (eV) is the kinetic energy gained by an elemental charge accelerated by a potential difference of 1 Volt.

1 eV = 1.602 x 10-19 J

keVeV

JVCVqU

6000,6

106.9)000,6)(1060.1( 1619

For a conservative force, the kinetic energy gained or lost is equal to the difference in potential energy:

[1] An electron acquires 7.45 x 10-17 J of kinetic energy when it is accelerated by an electric field from plate A to plate B. What is the potential difference and which plate is at the higher potential?

A B

E

KEΔW

qW

V C 10x 1.60

J 10x 7.4519

17

V 466V Plate B has a positive chargeand is at a higher potential.

E

q

x

x =1.00 cm q = 1.60 x 10-19 CE = 2000 N/C m = 9.1 x 10-31 kg

[2] a) Find the speed of the charge at the lower plate.

VΔqW

WKEΔ

VΔqqEx

ExVΔ m 0.01 N/C 2000VΔ

b) Find the potential through which the charge moves. qEx

2mv2

mqEx2

v

kg 10x 9.1

m 0.01N/C 2000C 10x 1.62v

13

19

V20VΔ m/s 10x 65.2v 6

m

x =1.00 cm q = 1.60 x 10-19 CE = 2000 N/C m = 9.1 x 10-31 kg

d) Find the acceleration of the charge.

qEF

mF

a

c) Find the force on the charge as it moves.

E

q

x

m

N/C 2000C 10x 1.6F 19

N 10x 2.3F 16

kg 10x .19

N 10x 2.313

16

ax2vv 2o

2

x2v

a2

m 01.02

m/s 10x 2.6526

241 m/s 10x 3.5a

241 m/s 10x 3.5a

or

Make sure that we understand the difference between Potential and Electric Potential

Energy:

V (in Volts) = Potential

a property of a certain position in an Electric Field with or without

charges placed there

E

-

EPE (in Joules) = Electric Potential Energy

a property of charges placed at a certain position in an external Electric Field

+

- E

d

E

V

dV

E

EdV

10 V 0 V5 V 2.5 V7.5 V

0 cm-10 cm -5 cm

Let E = 100 N/C d = 10 cm

m 0.10N/C 100V

V 10V

Using potentials instead of fields can make solving problems much easier – potential is a scalar quantity, whereas the field is a vector.

Electric Potential 17

Convention: V=0 at infinite r

potential) (electric r

QkV e

sd

Electric Field:

outward) (radially 2r

QkE e

abe

ab

rrQk

VVV

11

Electric Potential:

Electric potential at adistance r from a positive charge Q

Electric potential at adistance r from a negative charge Q

The electric potential due to a point charge

r

QkV

-V

+r-r

+V

+r-r

rkQ

V

r

QkV

rkQ

V

Electric Potential 17

For a system of point charges Qi at distances ri from a point P:

i i

ie r

QkV … an algebraic sum of scalars!

Q1 Q4

Q3 Q2

P

r1

r2

r3

r4

q r What is the Potential at this point?

V kqr

Notes:

1) Include the sign of q in your calculation! (+ or -)

3) The equation can also be used for a charged sphere:

+

++++ +

++

++

rV k

qr

Total charge

Distance from center

2) Potential Difference can also be calculated:

V = V2 – V1 kqr

kqr2 1

4) Electric Potential is a scalar not a vector

V = V1 + V2 + V3 + … (an

algebraic sum, not a vector sum)

i i

i

o r

qV

4

1

[4]What is the electric potential at a distance of 2.5 x 10-15 m away from a proton?

rQ

kV m 10x 2.5

C 10x 1.60/CNm 10x 8.99

15

19229

V 10x 8.5 5

+q

+q

-q -qP

dd

d

d

[5] Find the potential V at point P due to the four charges.

Web Link: Complex Electric Field

Ans: 13700 N/C, 100o from +ve x axis; -18,000 V; -0.072 J

[6] The +4 μC charge is located at 4 m on the x-axis and the -6 μC is located at +2 m on the y-axis as shown below.

a) Calculate the magnitude and determine the direction of the electric fields at the origin due to the +4 μC charge and due to the -6 μC charge.

b) Calculate the electric potential at the

origin.

c) Calculate the work done to bring a +4 μC from infinity to the origin.

Example: Approaching a Charged Sphere

A proton is fired from far away at a 1.0 mm diameter glass sphere that has a charge of q=+100 nC. What is the initial speed the proton must have to just reach the surface of the glass?

f f i iK U K U 1 22

0 0p sp

s

q qK m v

r

72 21.86 10 m/sp s s

p s p s

Kq q Keqv

m r m r

Example: Moving Through a Potential Difference

A proton with a speed of vi = 2.0x105 m/s enters a region of space where source charges have created an electric potential. What is the proton’s speed after it has moved through a potential difference of DV=100 V?

f f i iK qV K qV

f i i f iK K qV qV K q V

1 12 22 2f imv mv q V

2 521.44 10 m/sf i

ev v V

m

What is vf if the proton is replaced by an electron?

2

6

2(electron)

5.93 10 m/s

f ie

ev v V

m

Electrostatic Precipitator

• A strong electric field produces ionization of gases entering the device.

• Most of the tiny particulates present in the flue gas become negatively charged, and stick to the walls that are at a positive electric potential.

The electrostatic precipitator is highly effective in removing tiny particulates (e.g. carbon and metals) from the flue gases of coal-burning power plants.

Cathode-Ray Tube (CRT)

The electron beam is scannedin a raster pattern across the phosphors on the CRT screen.

– +

+

–

http://ap-physics.david-s.org