656_PartUniversity Physics Solution

# 656_PartUniversity Physics Solution - Gausss Law 22.25 22-7...

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Gauss’s Law 22-7 22.25. IDENTIFY: The magnitude of the electric field is constant at any given distance from the center because the charge density is uniform inside the sphere. We can use Gauss’s law to relate the field to the charge causing it. (a) SET UP: Gauss’s law tells us that 0 q EA = P , and the charge density is given by 3 (4/3) qq VR ρ π == . EXECUTE: Solving for q and substituting numbers gives 22 1 2 2 2 8 00 (4 ) (1750 N/C)(4 )(0.500 m) (8.85 10 C /N m ) 4.866 10 C qE A E r ππ −− = × = × PP . Using the formula for charge density we get () 8 73 3 3 2.60 10 C/m . 0.355 m × = = × (b) SET UP: Take a Gaussian surface of radius r = 0.200 m, concentric with the insulating sphere. The charge enclosed within this surface is 3 encl 4 3 qV r ρρ ⎛⎞ ⎜⎟ ⎝⎠ , and we can treat this charge as a point-charge, using Coulomb’s law encl 2 0 1 4 q E r = P . The charge beyond r = 0.200 m makes no contribution to the electric field. EXECUTE: First find the enclosed charge: 3 37 3 9 encl 44 2.60 10 C/m 0.200 m 8.70 10 C 33 qr ρπ × = × ⎢⎥ Now treat this charge as a point-charge and use Coulomb’s law to find the field: 9 92 2 3 2 9.00 10 N m /C 1.96 10 N/C 0.200 m E × =×⋅ = × EVALUATE: Outside this sphere, it behaves like a point-charge located at its center. Inside of it, at a distance r from the center, the field is due only to the charge between the center and r . 22.26. IDENTIFY: Apply Gauss’s law and conservation of charge. SET UP: Use a Gaussian surface that lies wholly within he conducting material. EXECUTE: (a) Positive charge is attracted to the inner surface of the conductor by the charge in the cavity. Its magnitude is the same as the cavity charge: nner 6.00 nC, i q =+ since 0 E = inside a conductor and a Gaussian surface that lies wholly within the conductor must enclose zero net charge. (b) On the outer surface the charge is a combination of the net charge on the conductor and the charge “left behind” when the 6.00 nC + moved to the inner surface: tot inner outer outer tot inner 5.00 nC 6.00 nC 1.00 nC. q q =+⇒= −= = EVALUATE: The electric field outside the conductor is due to the charge on its surface. 22.27. IDENTIFY: Apply Gauss’s law to each surface. SET UP: The field is zero within the plates. By symmetry the field is perpendicular to the plates outside the plates and can depend only on the distance from the plates. Flux into the enclosed volume is positive. EXECUTE: 23 and SS enclose no charge, so the flux is zero, and electric field outside the plates is zero. Between the plates, 4 S shows that EA q σ A − = and 0 . E σ = P EVALUATE: Our result for the field between the plates agrees with the result stated in Example 22.8. 22.28. IDENTIFY: Close to a finite sheet the field is the same as for an infinite sheet. Very far from a finite sheet the field is that of a point charge.

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## This note was uploaded on 07/16/2011 for the course PHY 2053 taught by Professor Buchler during the Spring '06 term at University of Florida.

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656_PartUniversity Physics Solution - Gausss Law 22.25 22-7...

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