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Chapter 15

Chapter 15. Electric Forces and Electric Fields. Electric Flux. Field lines penetrating an area A perpendicular to the field The product of EA is the flux, Φ In general: Φ E = E A cos θ. Electric Flux, cont. Φ E = E A cos θ

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Chapter 15

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  1. Chapter 15 Electric Forces and Electric Fields

  2. Electric Flux • Field lines penetrating an area A perpendicular to the field • The product of EA is the flux, Φ • In general: • ΦE = E A cos θ

  3. Electric Flux, cont. • ΦE = E A cos θ • The perpendicular to the area A is at an angle θ to the field • When the area is constructed such that a closed surface is formed, use the convention that flux lines passing into the interior of the volume arenegative and those passing out of the interior of the volume are positive

  4. Example 15.6 – page 518 • Calculate electric flux through a closed surface: Consider a uniform electric field oriented in the x-direction. Find the electric flux through each surface of a cube with edges L oriented as shown in the figure, and the net flux. • Strategy: use the definition of the electric flux; E and A=L2 are the same for all cube’s surfaces; the only difference is angle θ between electric field vector E and normal vector to the surface.

  5. Gauss, Karl Friedrich (1777-1855) • German mathematician who is sometimes • called the "prince of mathematics." He • was a prodigious child, at the age of three • informing his father of an arithmetical error • in a complicated payroll calculation and • stating the correct answer. • In school, when his teacher gave the • problem of summingthe integers from • 1 to 100 (an arithmetic series) to his • students to keep them busy, Gauss • immediately wrote down the correct • answer 5050 on his slate.

  6. Gauss’ Law • Gauss’ Law states that the electric flux through any closed surface is equal to the net charge Q inside the surface divided by εo • εo is the permittivity of free space and equals 8.85 x 10-12 C2/Nm2 • The area in Φ is an imaginary surface, a Gaussian surface, it does not have to coincide with the surface of a physical object

  7. Electric Field of a Charged Thin Spherical Shell • The calculation of the field outside the shell is identical to that of a point charge • The electric field inside the shell is zero

  8. Example 15.7 – page 520 • A spherical conducting shell of inner radius a and outer radius b carries a total charge +Q distributed on the surface of a conducting shell. The quantity Q is taken to be positive. • Find the electric field in the interior of the conducting shell, for r<a, and; • The electric field outside the shell, for r>b. • If an additional charge of –Q is placed at the center, find the electric field for r>b.

  9. Electric Field of a Nonconducting Plane Sheet of Charge • Use a cylindrical Gaussian surface • The flux through the ends is EA, there is no field through the curved part of the surface • The total charge is Q = σA • Note, the field is uniform

  10. Electric Field of a Nonconducting Plane Sheet of Charge, cont. • The field must be perpendicular to the sheet • The field is directed either toward or away from the sheet

  11. Parallel Plate Capacitor • The device consists of plates of positive and negative charge • The total electric field between the plates is given by • The field outside the plates is zero

  12. Example 15.8 – page 521 • Find the electric field above and below a nonconducting infinite plane sheet of charge with uniform positive charge per unit area σ (Figure 15.30, page 522). • Gauss’s law: • Electric flux comes from two ends, each having area A0.

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