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This resource covers essential concepts of electric potential, potential difference, and electrical work, focusing on the relationships between charge, voltage, and energy. Key computations include how to calculate potential differences and work done in moving charges through electric fields, utilizing both standard and electronvolt units. Through examples, it illustrates the transformation of electric potential energy into kinetic energy and provides formulas for practical conversions between energy measurements.
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Potential IV . Electric ______________ • Move some _______________ from point A to B in the E ( ________________ ) field of a charged object: charge q electric lines E field ________ A q charged object F B It takes ____________ to move q closer because it requires a ____________ applied for a ______________ . work W force F distance d
Because of the work done, the energy of the ___________of charges is stored up as an electric _______ . PE system The electric _______________________ V is defined as the work done per charge: potential difference V = W / q no direction scalar V is a _____________ because it has ___________________ units of V: By definition, one _______________________ = 1 _________ or 1 _______ = 1 _____ [V] = [ ] / [ ] q W [V] = / J C volt joule/coulomb V J/C Because of its units, the potential difference is often called the ________________ . voltage
Ex 1: 18 J of work are required to move 2.0 C of charge between two points in an electric field. Find the electric potential difference between the two points. Given: V = W / q W 18 J = V = 18 J /2.0 C = 9.0 J/C = 9.0 V 2.0 C = q V = ? How much work would be required to move 4.0 C of charge between the same two points of the problem above? V = W / q = = q = 4.0 C 9.0 V V = W/ 4.0 C 9.0 V 36 J W W = ? How much potential energy is stored up as a result? 36 J
NOTE: V = W/q can be used with 2 sets of units! • When working with small amounts of charge and energy, you often use _______________ units. • For W, instead of joules, use units: • ______________________ • For q, instead of coulombs, use units: • the number of ______________________ • Using these new units, the potential difference V in: • V = W/q will have units: [ V ] = [ ] / [ ] • which again be volts: = / • You get V because _________________ . • smaller • electronvolts, eV elementary charges, e • q • W • V • eV • e the e's cancel
Ex: 20 electronvolts of work are required to move 8 electrons between two points in an electric field. Find the electric potential difference between the two points V = W / q = 20 eV / 8 e = 2.5 eV/e = 2.5 V Given: 20 eV = W 8 e = q V = ? Ex: How much work (in eV) is done on a proton when it is accelerated through a potential difference of 68,000 volts? V = W / q = / W = W 1 e 68,000 eV 68,000 V How much kinetic energy will the proton have as a result? DKE = W = 68,000 eV
Conversions between units. See PhysRT, page ____: 1 electronvolt (eV) = _____________________ J Notice that this number is the same as the one in the conversion: 1 elementary charge (e) = _____________________ C 1 Ex. Convert 9.6 x 10-19 J to eV. 9.6 x 10-19 J x ___________________ = ______________ eV Ex. Convert 4640 eV to J. 4640 eV x ___________________ = ___________________ J
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Conversions between units. See PhysRT, page ____: 1 electronvolt (eV) = _____________________ J Notice that this number is the same as the one in the conversion: 1 elementary charge (e) = _____________________ C 1 1.60 x 10-19 1.60 x 10-19 Ex. Convert 9.6 x 10-19 J to eV. 1 eV 1.60 x 10-19 J 6.0 9.6 x 10-19 J x ___________________ = ______________ eV Ex. Convert 4640 eV to J. 1.60 x 10-19 J 1 eV 7.42 x 10-16 4640 eV x ___________________ = ___________________ J
Summary of electric potential units: eV J e C V V eV's are NOT a unit of __________________________ . They are a unit for __________ or _____________. Also, you cannot mix: _____ with _______ or: _____ with _______ One of them must be converted first! potential difference work energy e J eV C
Ex. Find the work done in joules in moving 5 electrons through a potential difference of 6.0 V. Solution 1: Find answer in eV, then convert: V = W /q = = W ________ eV x _______________ = ____________ J Solution 2: Convert e to C first, then calculate. _______ e x __________________ = ____________ C V = W /q = = W 5 e 6.0 V W / 30 eV 1.60 x 10-19 J 1 eV 4.8 x 10-18 30 1.60 x 10-19 C 1 e 8.0 x 10-19 5 8.0 x 10-19 C 6.0 V W / 4.8 x 10-18 J
NOTES: • Electric potential difference represents the ___________ • per ____________ stored up in the system of charges. PE charge • If the charges are allowed to move freely through a • potential difference, they will gain the energy as _____ • in the same way that a ___________ object gains it. KE falling gravity: electricity: PE = 100 J KE = 100 J PE = 100 J KE = 100 J
3. A van der Graff is a device that produces a strong _____________ field. This provides a large _____________ __________________ to do ______________ on charges: V = W/q W = As a result, the _______ of the charges is increased: This increases their _____________ because: Neutrons cannot be accelerated by van der Graff's because they are ________________ and are unaffected by ______________ fields. potential electric work difference qV KE W = DE = DKE speed KE = (1/2) mv2 neutral electric
