1. Voltage

2. Volt  The electric potential is related to the potential energy. Compare to test charge  The unit of electric potential is the volt (V). 1 V = 1 J/C q q d F E VV

3. Field Measure  The electric field is most commonly measured in V/m. Show that this is consistent with a measurement in N/C.  Use the definitions of N and J to link the two definitions of electric field. 1 N/C = 1 (kg m / s 2 ) / C 1 N/C = 1 (kg m) / (s 2 C) 1 V/m = 1 (J/C) / m 1 V/m = 1 (N m / C) / m=1 N/C 1 V/m = 1 (kg m 2 / s 2 ) / (C m) 1 V/m = 1 (kg m) / (s 2 C)

4. Electric Work  A cathode ray tube accelerates electrons across a potential of 20 kV. Find the speed of the electrons at the screen.  The potential can be converted to an energy. qV = (1.6 x 10 -19 C)(2 x 10 4 V) = 3.2 x 10 -15 J.  The potential energy becomes kinetic energy. qV = ½ mv 2  Solve for the speed v. v = 8.4 x 10 7 m/s

5. Uniform Field  A uniform electrical field has the same magnitude and direction at all points.  A charge moving parallel to the field lines changes potential by V = Ed.  A charge moving perpendicular to the field lines has no change in potential. q q d F E UU

6. Conductor Potential  There is no field within a conductor. External field neutralized by polarizationExternal field neutralized by polarization  A test charge at one end moved to the other end would not change potential.  All points on a conductor are at the same potential. qq

7. Voltage Source  A voltage source is called a battery.  A battery attached to a conducting plate places the same potential across the plates.  A uniform electric field exists between the plates. 

8. Internal Field  A 12-V source is connected to parallel plates 2.0 mm apart. Find the magnitude of the field between the plates.  The relationship between the field and voltage is V = Ed. Solve for E = V/dSolve for E = V/d  The field is E = (12 V) / (0.002 m) = 6.0 kV/m.  next