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APPLIED PHYSICS AND CHEMISTRY ELECTRICITY LECTURE 4 Work and Electric Potential

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Work Remember: Work is using a force to move something a distance W = Fd For an electric charge: Work is done when a force moves a test charge farther away from a negative source charge We are increasing the potential energy of the test charge Work is proportional to the magnitude of the charge moved Because electric force is related to the magnitude of the charges involved! Vb – Va = W/q

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Electric Potential Difference in electric potential measures the effect of location within the electric field Low potential: High potential: Only changes in electric potential are related to work done! Electric potential is measured in Volts 1 volt = 1 Joule/Coulomb ++ ++

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Uniform Electric Fields Uniform fields can be created using parallel flat plates with charges Work done to move a charge in this field is Fd V = Fd/q (potential is work/charge) Since F/q equals E (field intensity): V = Ed

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Problem Two parallel plates are 0.500 m apart. The electric field intensity between them is 6.00 x 10 3 N/C. What is the potential difference between the plates? What we know: D = 0.500 m E = 6.00 x 10 3 N/C Equation: V = Ed Substitute: V = (6.00 x 10 3 N/C)(0.500 m) Math: V = 3.00 x 10 3 Nm/C = 3.00 x 10 3 J/C

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Problem Continued Two parallel plates are 0.500 m apart. The electric field intensity between them is 6.00 x 10 3 N/C. What work is done moving a charge equal to that on one electron from one plate to the other? What we know: Q = 1.6 x 10 -19 CV = 3.00 x 10 3 V Equation: V = W/q so W = Vq Substitute: W = (3.00 x 10 3 V)(1.6 x 10 -19 C) Math: W = 4.8 x 10 -16 CV = 4.8 x 10 -16 J

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Another Problem A voltmeter measures the potential difference between two parallel plates to be 60.0 V. The plates are 0.030 m apart. What is the magnitude of the electric field intensity? What we know: V = 60.0 V d = 0.030 m Equation: V = Ed so E = V/d Substitute: E = (60.0 V)/(0.030 m) Math: E = 2.0 x 10 3 V/m = 2.0 x 10 3 N/C

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