Finding Charge from Gauss's Law - SFSU Physics & Astronomy
Finding Charge from Gauss’s Law • E = 2(N/C)r everywhere on surface of a 10m diameter sphere centered on origin. What is the net charge in the sphere?
r Φ E = ∫ E • nˆ dA
R=n
= ∫ (2( N / C )rˆ • rˆdA = 2 N / C ∫ dA
Qin = (2 N / C )4πR = 200πN • m / C = ε0 2
• Qin=5.6 x 10-9 C
2
Finding E from known charge distribution • A few point charges – Use Coulomb law and superposition • Continuous charge distribution with symmetry – Use Gauss’s law – May work when very close to finite object • General case of charge distribution – Integrate Coulomb law over charge distribution
r kdq E = ∫ 2 rˆ r
Finding E with Gauss’s Law - Steps • Draw sample field lines. Choose coordinate system so E has only 1 component at each point: – Cartesian: x, y, z components – Spherical: radial, azimuthal, polar components – Cylindrical: radial, axial, azimuthal components • Construct (imaginary) “Gaussian surface” through field point so that all parts of surface have E n or E n. (Need symmetric charge distribution)
r ∫ E • nˆdA = E∏ A = Qin / ε 0
• E=(Qin/ε0A). Should already know direction of E. Q1
Example: Uniformly charged string (filament) • If field point is close to string but far from ends of string, approximate string as infinitely long P (field point)
Gaussian surface (cylinder) • Field approx. symmetric at Gaussian surface • Case lacking symmetry (must integrate Coulomb): P
Filament has linear charge density λ Charge in length dl is dQ= λdl Charge in length is λh Cylindrical coordinate system E is radial Gaussian surface
Follow steps to get E from Gauss’s law • Draw sample field lines. Use cylindrical coordinates. φ r z • Gaussian surface: cylinder of radius r, height h n = r on curved surface, so E • n = E there. n ⊥ E on flat faces, so E • n = 0 there
r ∫ E • nˆdA = E (2πrh)
Qin λh λ 2πrhE = ⇒E= = ε0 ε 0 2πrh 2πε 0 r
• Thus
r E=
λ rˆ 2πε 0 r
• Note that field falls off as 1/r ; compare to pt. charge • Question: What is the E field inside spherical shell of charge? Charge q uniformly distributed through shell • Question 2 • Question 3 • If we have a short filament, must use direct integration
Field at distance y above filament of length l • Consider pairs of differential charge elements dQ=λdx located at +x and -x (similar to previous case of pair of point charges), then integrate over x
r dE = 2dE1, y ˆj = 2dE1 cosθˆj kdQ1 y ˆ 2kyλdx ˆ j j= =2 2 3 r r r r 2kyλdx ˆ l/2 E[ y ] = ∫x = 0 j 2 2 3/ 2 x +y kQ ˆj = 2 2 y y + (l / 4)
(
)
r Φ E = ∫ E • nˆ dA
R=n
= ∫ (2( N / C )rˆ • rˆdA = 2 N / C ∫ dA
Qin = (2 N / C )4πR = 200πN • m / C = ε0 2
• Qin=5.6 x 10-9 C
2
Finding E from known charge distribution • A few point charges – Use Coulomb law and superposition • Continuous charge distribution with symmetry – Use Gauss’s law – May work when very close to finite object • General case of charge distribution – Integrate Coulomb law over charge distribution
r kdq E = ∫ 2 rˆ r
Finding E with Gauss’s Law - Steps • Draw sample field lines. Choose coordinate system so E has only 1 component at each point: – Cartesian: x, y, z components – Spherical: radial, azimuthal, polar components – Cylindrical: radial, axial, azimuthal components • Construct (imaginary) “Gaussian surface” through field point so that all parts of surface have E n or E n. (Need symmetric charge distribution)
r ∫ E • nˆdA = E∏ A = Qin / ε 0
• E=(Qin/ε0A). Should already know direction of E. Q1
Example: Uniformly charged string (filament) • If field point is close to string but far from ends of string, approximate string as infinitely long P (field point)
Gaussian surface (cylinder) • Field approx. symmetric at Gaussian surface • Case lacking symmetry (must integrate Coulomb): P
Filament has linear charge density λ Charge in length dl is dQ= λdl Charge in length is λh Cylindrical coordinate system E is radial Gaussian surface
Follow steps to get E from Gauss’s law • Draw sample field lines. Use cylindrical coordinates. φ r z • Gaussian surface: cylinder of radius r, height h n = r on curved surface, so E • n = E there. n ⊥ E on flat faces, so E • n = 0 there
r ∫ E • nˆdA = E (2πrh)
Qin λh λ 2πrhE = ⇒E= = ε0 ε 0 2πrh 2πε 0 r
• Thus
r E=
λ rˆ 2πε 0 r
• Note that field falls off as 1/r ; compare to pt. charge • Question: What is the E field inside spherical shell of charge? Charge q uniformly distributed through shell • Question 2 • Question 3 • If we have a short filament, must use direct integration
Field at distance y above filament of length l • Consider pairs of differential charge elements dQ=λdx located at +x and -x (similar to previous case of pair of point charges), then integrate over x
r dE = 2dE1, y ˆj = 2dE1 cosθˆj kdQ1 y ˆ 2kyλdx ˆ j j= =2 2 3 r r r r 2kyλdx ˆ l/2 E[ y ] = ∫x = 0 j 2 2 3/ 2 x +y kQ ˆj = 2 2 y y + (l / 4)
(
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