FIELDS

1 / 41

# FIELDS - PowerPoint PPT Presentation

FIELDS.  V. Gradient of slope =.  x. Electrical field. +ve charge moving towards –ve charge. Higher potential. +. V. x. Lower Potential. Increases if V bigger and  x smaller. - Potential gradient = Electrical Field Strength. +. -. +. Non-uniform fields. Uniform field.

I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.

## PowerPoint Slideshow about 'FIELDS' - vic

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript

### FIELDS

V

x

Electrical field

+ve charge moving towards –ve charge

Higher potential

+

V

x

Lower Potential

Increases if V bigger and x smaller

- Potential gradient = Electrical Field Strength

+

-

+

Non-uniform fields

Uniform field

Field Strength Line patterns

(line to indicate direction of force felt by a +ve charge)

-

+

Field strength lines

Equipotential, 90 to field strength lines

800V

0V

200V

400V

600V

Lines of equipotential – no work done on charge

Equipotentials in a uniform field will be equally spaced

Describing a uniform field

Electric Force

• Force between two charges
• May be attractive (-ve) or repulsive (+ve)

Symbol = F

Units = N

E = F

q

E = V

d

distance between charges (plates)

Describing a uniform field

Electric Field Strength

• Force per unit +ve charge

Important due to repulsive/attractive nature of force

Symbol = E

Unit = N C-1

or V m-1

Describing a uniform field

Electric Potential Energy

• Energy of a +ve charged particle (q) due its position in an electric field

Symbol = EPE

Units = C V = J

EPE = qV

EPE = qV

10V

+

+

0V

EPE = 0

Acceleration due to electric field

 potential energy =  kinetic energy

qV = ½mv2

Important point:

K.E. gained is the same for both particles

Velocity is different due to difference in mass

Electron velocity quickly approaches the speed of light - relativistic.

Velocity equation only valid when v << c

The electron volt (eV)

A measure of energy

qV = ½mv2

1 eV = Energy gained when an electron is subjected to a potential difference of 1V

1 eV = 1.6 x 10-19C x 1V = 1.6 x 10-19 J

Millikan’s Oil Drop Experiment

to measure charge on an electron

Forces experienced by oil drop?

+

F (electric attraction) = qE

-

Oil drop (charge q)

F (weight) = mg

-

If the oil drop hovers then…

mg = qE

The cunning bit...

Then ionise oil drop (using radioactive source)

Re-adjust the voltage (and therefore the size of E) to make oil drop hover again.

Change in voltage proportional to charge on an electron

Apple and moon experiment

by Newton

Question - how does gravity extend into space?

Newton calculated...

Acceleration of Moon towards Earth

Acceleration of an apple towards Earth

Conclusion...

Angular acceleration of moon towards earth (due to circular motion)

Strength of earth’s gravity at 60RE

=

Gravity changes at a rate of inverse square of distance

This extended gravitational force out into the universe - an amazing result (!)

F = -Gm1m2

r2

Gravitational force, F

F = gravitational force, N

G = gravitational constant, 6.67 x 10-11 N m2 kg-2

m1 = mass of first object, kg

m2 = mass of second object, kg

r = distance between the two objects, m

90 to velocity of object

Needed for circular motion

Gm1m2

m2v2

=

r

r2

Requirement for an object to orbit

Fgravitational = Fcentripetal

N.B. m1 = mass at centre of orbit, m2 = mass of satellite

Very important

Gm1m2

m242r

=

r2

T2

T2

42r3

=

Gm1

v = 2r

Remembering

T

(Kepler III)

The satellite must be travelling fast enough for its orbit radius (Kepler III)

• Not faster enough - orbit will collapse
• Too fast - will overcome gravitational forces and escape

Relative to the earth it doesn’t move

T =

Orbits N-S (over the poles), the earth rotates and so it looks at a different place each orbit.

T =

Types of orbit

Geostationary

24 hours

Polar

90 minutes

Gravitational field strength, g

- better known as gravity

If there is a force there is an acceleration

F = ma

If the force is due to gravitational forces then acceleration is acceleration due to gravity

F = mg

F = -GMm

r2

= -GM

mr2

r2

g = F

units N kg-1

m

Or

In words, gravitational force per unit mass acting at a point

Know

g = -GMm

Then

Remember:

work = force x distance (in direction of force)

= mg x h

Uniform Field (near surface of Earth)

= mgh

Gravitational Potential Energy

Stored ability to do work

- something else has done work to get the object to that point

mg(h + 3Δh)

mg(h + 2Δh)

mg(h + Δh)

Δh

mgh

h

h=0

Change in GPE

Area = mg Δh

= work done

Δh

h1

h2

Work done moving an object by Δh (near the earth’s surface, g  constant)

Force (mg) /N

Height /m

V = mgh = gh

Symbol = V

m

Unit = J kg-1

Gravitational potential

...“The work done to move unit mass from infinity to that point”...

Don’t forgeth  r

Gravitational potential is the total work, against the gravitational force, for 1kg to go from a point where g = 0 to the point in question where g = x N kg-1.

g = 0 N kg-1 at r = 

g = 9.8 N kg-1 at r = 6.4 x 106m

Space calculations

GPE = 0

GPE = - x

A convention...

Earth only calculations

GPE = x

GPE = 0

r /m

Δr

0

r

= m g dr

r

Force (mg) / N

Work done moving an object from  to r (Δr)

Area = work done

N.B. g isn’t constant (non-uniform field)

g dr

V =

r

V = - GM dr

= GM

= - GM

r

r

r2

r

r

Gravitational potential - work done on unit mass i.e. m = 1kg

g = - dV

dr

Another interesting point...

g dr

V =

r

Can be rearranged to ...

The gradient of gravitational potential is gravitational field strength.

F = - GMm

Gravitational force

r2

g = - GM

Gravitational field strength

r2

V = - GM

Gravitational potential

r

Energy conservation

Etotal = Ekinetic + Epotential

mgh = Etotal - ½mv2

gh = constant - ½v2

V  -½v2

r

r

-½v2

V

The gradient of either graph is g

Energy gained if falling into hole

Escape velocity

V = 0, r = 

V = -62.5MJ kg-1

Energy required to get out of hole

v   -2V

V is -ve

Stationary object (v = 0),

at V = 0

½mv2 + mV = ETotal

0 + 0 = 0

Nudge object into well, ETotal = 0 K.E. increases as P.E. become more -ve

½mv2 + mV = 0

At Earth’s surface V = -62.5MJ kg-1, a 1kg mass will hit the ground at ~11km s-1 if nudged into well.

Conversely...

A 1kg mass launched at 11km s-1 will just make it to V = 0, the brim of the potential well.

11km s-1 = escape velocity