Electrostatics

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# Electrostatics - PowerPoint PPT Presentation

Electrostatics. +. +. -. -. -. +. Conservation of Charge. Charge can neither be created nor destroyed. Positive ions ---- fewer electrons than protons. Negative ions ---- fewer protons than electrons. Electric Charge is measured in Coulombs

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Presentation Transcript
Conservation of Charge

Charge can neither be created nor destroyed

Positive ions ---- fewer electrons than protons

Negative ions ---- fewer protons than electrons

Electric Charge is measured in Coulombs

6.3x1018 electrons make -1.0 C of charge

6.3x1018 protons make +1.0 C of charge

Conservation of Charge

Charge can neither be created nor destroyed

rub electrons from a bar with fur

bar becomes positively charge by

the exact amount that fur becomes

negatively charged.

bar becomes a tiny bit less massive

Coulomb’s Law

The interaction force between two charges is:

directly proportional to the size of each charge (q1 and q2)

and

inversely proportional to the square of their separation

distance (d)

k= 9.0 x 109 N/m2/C2 Fe>>>Fg

Coulomb’s Law

The interaction force between two charges is:

directly proportional to the size of each charge (q1 and q2)

Double either q1 or q2 then F doubles.

Double both then F quadruples

Coulomb’s Law

The interaction force between two charges is:

inversely proportional to the square of their separation

distance (d)

Double the separation distance then F is reduced to (1/4)

Halve the separation distance then F is quadrupled (4x)

Coulomb’s Law

The interaction force between two charges is:

inversely proportional to the square of their separation

distance (d)

triple the separation distance then F is reduced to (1/9)

(1/3) the speration distance then F is increased 9 fold (9x)

Coulomb’s Law

The interaction force between two charges is:

inversely proportional to the square of their separation

distance (d)

If separation distance is increased by 10 then F

(Reduces/increases) by _________________

Coulomb’s Law

+

+

As the charges above are released the force on each

(increases or decreases)

Coulomb’s Law

+

+

As the charges above are released the speed of each

(increases or decreases)

Coulomb’s Law

+

+

As the charges above are released the acceleration of each

(increases or decreases)

Coulomb’s Law

?

+

As the charges above are released the speed of each

increases. Thus the green object has a __________

charge. Positive, negative, can’t tell

Coulomb’s Law

?

+

As the charges above are released the force on each

increases. Thus the green object has a __________

charge. Positive, negative, can’t tell

Conductors

Conductors have very loosely bound electrons. That is electrons that are not really attached to one particular nucleus. These electrons are sometimes called free electrons because they move freely when exposed to an electric field

Gold

Copper

Silver

Ionic solutions (salt water)

Insulators

Insulators have very tightly bound electrons. That is electrons that are firmly attached to one particular nucleus. These electrons are very hard to set in motion throughout the material

Glass

Dry wood

Plastic

Semiconductors

Semiconductors have moderately bound electrons. These electrons can be set into motion throughout the material when a moderately strong electric field is established in the material.

Carbon

Silicon

Superconductors

Superconductors have no electrical resistance to charge flow (infinite electrical conductivity)

Very cold silver (-269 °C)

Charging

Friction

Contact

Induction

Charge Polarization

+ -

- -

- -

- -

- -

- -

- -

+ -

- -

- -

- -

- -

- -

- -

+ -

+ -

F

Electric Field

E=F/q

or

F=q E

-

-

-

-

-

-

+

+

+

+

+

+

F

+

q

Uniform Electric

Field between

two charged plates

Electric Potential (Volts)

-

-

-

-

-

-

+

+

+

+

+

+

Electric Potential energy (J)

Charge (C)

EP = EPE / q

Volt=Joule/Coulomb

F

+

q

Electric Potential Energy = Work

Electric Potential Energy = Charge x Volts

1 Joule= Coulomb x Volt

Electric Potential (Volts)

-

-

-

-

-

-

+

+

+

+

+

+

Electric Potential energy (J)

Charge (C)

F

+

q

What is the electric

potential between two

plates when it takes 2.0 J

of work to move a 0.001 C

charge from - to + plate?

What is the electric potential between two plates when it takes 2.0 J of work to move a 0.001 Ccharge from - to + plate?

Given: W=EPE=2.0 J Charge=0.001 C

Want: EP

Solution: Electric Potential=

Electric Potential energy (J)

Charge (C)

Given: W=EPE=2.0 J Charge=0.001 C

Want: EP

Solution: Electric Potential=

Electric Potential energy (J)

Charge (C)

=2.0 J/0.001 C = 2000 Volts

1 Volt=1J/C