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Electric Potential Difference and Work

Point A is located near charge +Q. The electric potential at A is V A. Electric Potential Difference and Work. To bring a charge q from infinity to point A, we must do WORK on q:. A. +q. +Q. Work =  E E. Work = E EA - E E. Work = qV A - 0. Work = qV A.

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Electric Potential Difference and Work

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  1. Point A is located near charge +Q. The electric potential at A is VA Electric Potential Difference and Work To bring a charge q from infinity to point A, we must do WORK on q: A +q +Q Work =  EE Work = EEA- EE Work = qVA- 0 Work = qVA

  2. We can define the units of electric potential in this way: The electric potential at a point in an electric field is 1 Volt if 1 Joule of work is done to bring 1 Coulomb of charge to that point from infinity. Electric Potential Difference, V: This is the difference in electric potential between two points in an electric field. VB VA Potential difference between A and B: V = VB-VA +Q Imagine now moving charge q from A to B. The work done and change in electric potential energy is: W= q V= q(VB-VA) = EE

  3. + - Potential difference decreases in the direction of the electric field  +q -a positive test charge LOSES electric potential energy as it moves from the positive to negative plate Example: A small sphere with charge -3.0 C creates an electric field. Find the potential difference between points A and B Which point is at a higher potential? Would a positive test charge gain or lose energy in moving from A to B? -3.0 C 2.0 cm A 5.0 cm B

  4. Charged Parallel Plates B +  V d  A - To move q from A to B: Work done=  EE = q  V ----(1) Work done= FE*d =q*d -----------(2) Equating 1 and 2: This gives us an expression for the Electric potential difference (V) between two charged plates, given the electric field (N/C) between the plates and the plate separation (m). q  V = q d  V =  d

  5. It also allows us to determine the electric field, given the potential difference,  V, and the plate separation, d: =  V d 20 V B This expression also shows us that the potential difference between the plates is proportional to the distance: 15V  V - 10V 5V A 0V Ex. 2: Calculate the work done to move a 15 nC charge from the negative plate to the positive plate of a charged parallel plate system. The electric field strength is 25.0 N/C and the plates are separated by 8.0 cm. W= q  V = q d = (15 x 10 -9 C) ( 25.0 N/C)(0.080 m) = 3.0 x 10 -8 J

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