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Regents Physics Electric Fields

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Regents Physics. Electric Fields. Electric field. The region around a charged particle through which a force is exerted on another charged particle. Electric Field Line. The imaginary line along which a positive test charge would move in an electric field. - PowerPoint PPT Presentation

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Page 1: Regents Physics

Regents Physics

Electric Fields

Page 2: Regents Physics

Electric field

The region around a charged particle through which a force is exerted on another charged particle.

Page 3: Regents Physics

Electric Field Line

The imaginary line along which a positive test charge would move in an electric field

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Page 4: Regents Physics

The direction of an electric field is the direction of the force on a stationary positive test charge located at any point on a field line.

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Page 5: Regents Physics

On a curved field line, the direction of the field at any point is the tangent drawn to the field line at that point.

Electric field lines begin on positive charges (or at infinity) and end on negative charges (or infinity).

Page 6: Regents Physics

Field lines NEVER intersect.

Page 7: Regents Physics

Electric field strength

The force on a stationary positive test charge per unit charge in an electric field.

E = Fe/q

Fe=electrostatic force in newtons q=charge in coulombs

E=electric field strength in newtons per coulomb

VECTOR quantity!!!

Page 8: Regents Physics

Ex.

What is the magnitude of the electric field strength at a point in a field where an electron experiences a 1.0-newton force?

Page 9: Regents Physics

Ex.

What is the magnitude of the electric field strength at a point in a field where an electron experiences a 1.0-newton force?

Use E = Fe/q.

Fe=1.0N

Q=1.60 x 10^-19C

Page 10: Regents Physics

Ex.

What is the magnitude of the electric field strength at a point in a field where an electron experiences a 1.0-newton force?

Use E = Fe/q.

Fe=1.0N

Q=1.60 x 10^-19C = 6.3 x 10^18 N/C

Page 11: Regents Physics

Field Around a Point Charge or Sphere

Field lines extend radially outward from a positive point charge and radially inward from a negative point charge.

On a sphere, charge is distributed uniformly, and electric field lines are perpendicular to the surface.

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Page 12: Regents Physics

Sphere…

Electric field strength within a hollow, charged, conducting sphere is zero.

Page 13: Regents Physics

Field between two oppositely charged parallel plates

If d between two is relatively small compared to area, electric field is uniform.

Parallel to each other.

Electric field strength is same at every point between plates.

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Page 14: Regents Physics

Field between two oppositely charged parallel plates

Magnitude of electric force on an electron or a proton located at any point between two given oppositely charged parallel plates is the same.

Particle speed increasesas it approaches plates ofopposite sign.

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Page 15: Regents Physics

Potential Difference

The work done (or change in potential energy) per unit charge as a charged particle is moved between two points in an electric field.

V= W/q

Unit= joule/coulomb = volt

Page 16: Regents Physics

11/7/2008

Page 17: Regents Physics

volt

Derived SI unit for potential difference

Page 18: Regents Physics

If an elementary charge is moved against an electric field through a potential difference of one volt, the work done on the charge is:

W=Vq =(1.00V)(1.60 x 10^-19J)

=1.60x10^-19J (this gain in potential energy is called the electronvolt, eV)

Page 19: Regents Physics

Ex.

Moving a point charge of 3.2 x 10^-19 coulomb between points A and B in an electric field requires 4.8 x 10^-18 joules of energy. What is the potential difference between these points?

Page 20: Regents Physics

Ex.Moving a point charge of 3.2 x 10^-19 coulomb between points A and B in an electric field requires 4.8 x 10^-18 joules of energy. What is the potential difference between these points?

V=W/q

4.8x10^-18J / 3.2x10^-19C =

Page 21: Regents Physics

Ex.Moving a point charge of 3.2 x 10^-19 coulomb between points A and B in an electric field requires 4.8 x 10^-18 joules of energy. What is the potential difference between these points?

V=W/q

4.8x10^-18J / 3.2x10^-19C = 15V

Page 22: Regents Physics