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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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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 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 + q +
Electric Field E=F/q - + q
Electric Field E=F/q or F=q E - - - - - - + + + + + + F + q Uniform Electric Field between two charged plates
Electric Shielding E=0 inside metals - - - - - - + + + + + +
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
Capacitors & Energy Storage - - - - - - + + + + + + F + q