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Chapter S24

# Chapter S24 - From Chapter 23 Coulombs Law We found the...

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From Chapter 23 – Coulomb’s Law We found the fields in the vicinity of continuous charge distributions by integration: 2k E x λ = Line of Charge: o E 2 σ = ε Charged Plate: + + + + dE r x r R Are you interested in learning an easier way?

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Electric flux The field is uniform The plane is perpendicular to the field E E A Φ r
Flux for a uniform electric field passing through an arbitrary plane The field is uniform The plane is not perpendicular to the field E E A E Acos Φ = = θ r r ˆ n ˆ A An = r E E A Φ = r r g

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Rules for drawing field lines The electric field, , is tangent to the field lines. The number of lines leaving/entering a charge is proportional to the charge. The number of lines passing through a unit area normal to the lines is proportional to the strength of the field in that region. Field lines must begin on positive charges (or from infinity) and end on negative charges (or at infinity). The test charge is positive by convention. No two field lines can cross. E r # of electric field lines E Area r E N E A = Φ r
ˆ n A E r θ E r General flux definition ( 29 E ˆ E n A E Acos ∆Φ = = θ r r g The field is not uniform The surface is not perpendicular to the field A If the surface is made up of a mosaic of N little surfaces ( 29 N E i i i i 1 ˆ E n A = Φ = r g E ˆ E ndA E dA Φ = = r r r g g

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Electric Flux, General In the more general case, look at a small area element In general, this becomes cos E i i i i i E ∆Φ = = ×∆ E A 0 surface lim i E i i A E A d Φ = ×∆ = × E A
Electric Flux Calculations The surface integral means the integral must be evaluated over the surface in question In general, the value of the flux will depend both on the field pattern and on the surface The units of electric flux will be N . m 2 /C 2

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Electric Flux, Closed Surface The vectors Δ A i point in different directions At each point, they are perpendicular to the surface By convention, they point outward
Closed surfaces Q – 3 Q + +Q E ˆ E ndA E dA Φ = = r r r g g Ñ Ñ ˆ n or A point in the direction outward from the closed surface r Negative Flux Zero Flux

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Flux from a point charge through a closed sphere E ˆ E ndA E dA Φ = = r r r g g Ñ Ñ ˆ ˆ E n E n E = r r r P g 2 kq ˆ E r E constant r = = r r E sphere ˆ E ndA E dA E A Φ = = = r r r r g Ñ Ñ 2 E 2 o kq q 4 r 4 kq r Φ = π = π = ε
Gauss’s Law Gauss asserts that the proceeding calculation for the flux from a point charge is true for any charge distribution!!!

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