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ElectrostaticsElectrostatics Conservation of Charge Charge can neither be created nor destroyed Positive ions ---- fewer electrons than protons.

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ElectrostaticsElectrostatics

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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 6.3x10 18 electrons make -1.0 C of charge 6.3x10 18 protons make +1.0 C of charge

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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

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Coulombs 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 10 9 N/m 2 /C 2 F e >>>F g

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Coulombs 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

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Coulombs 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)

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Coulombs 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)

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Coulombs 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 _________________

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Coulombs Law As the charges above are released the force on each (increases or decreases) + +

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Coulombs Law As the charges above are released the speed of each (increases or decreases) + +

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Coulombs Law As the charges above are released the acceleration of each (increases or decreases) + +

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Coulombs Law As the charges above are released the speed of each increases. Thus the green object has a __________ charge. Positive, negative, cant tell ? +

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Coulombs Law As the charges above are released the force on each increases. Thus the green object has a __________ charge. Positive, negative, cant tell ? +

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ConductorsConductors 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)

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InsulatorsInsulators 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

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SemiconductorsSemiconductors 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

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SuperconductorsSuperconductors Superconductors have no electrical resistance to charge flow (infinite electrical conductivity) Very cold silver (-269 °C)

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ChargingCharging Friction Contact Induction

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Charge Polarization F

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Electric Field + + E=F/q q

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Electric Field - + E=F/q q

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Electric Field E=F/q or F=q E Uniform Electric Field between two charged plates + q F

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Electric Shielding E=0 inside metals

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Electric Potential (Volts) Electric Potential energy (J) Charge (C) q F EP = EPE / q Volt=Joule/Coulomb Electric Potential energy = Work

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Electric Potential (Volts) Electric Potential energy (J) Charge (C) q F What is the electric potential between two plates when it takes 2.0 J of work to move a C charge from - to + plate?

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Electric Potential energy (J) Charge (C) What is the electric potential between two plates when it takes 2.0 J of work to move a Ccharge from - to + plate? Given: W=EPE=2.0 J Charge=0.001 C Want: EP Solution: Electric Potential=

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Electric Potential energy (J) Charge (C) =2.0 J/0.001 C = 2000 Volts 1 Volt=1J/C Given: W=EPE=2.0 J Charge=0.001 C Want: EP Solution: Electric Potential=

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Capacitors & Energy Storage q F

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