# Voltage - PowerPoint PPT Presentation

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Voltage. 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. Volt. D V. d. E. F. The electric field is most commonly measured in V/m. Show that this is consistent with a measurement in N/C.

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Voltage

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## Voltage

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

### Volt

DV

d

E

F

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 / s2 ) / C

1 N/C = 1 (kg m) / (s2 C)

1 V/m = 1 (J/C) / m

1 V/m = 1 (N m / C) / m=1 N/C

1 V/m = 1 (kg m2 / s2) / (C m)

1 V/m = 1 (kg m) / (s2 C)

### Field Measure

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 104 V) = 3.2 x 10-15 J.

The potential energy becomes kinetic energy.

qV = ½ mv2

Solve for the speed v.

v = 8.4 x 107 m/s

### Electric Work

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

### Uniform Field

DU

d

E

F

There is no field within a conductor.

External 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.

q

q

### Conductor Potential

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.

### Voltage Source

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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/d

The field is E =

(12 V) / (0.002 m) = 6.0 kV/m.

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