22-1GAUSS’S LAW22.1. (a) IDENTIFYand SET UP: cos,EEdAφΦ=∫where φis the angle between the normal to the sheet ˆnand the electric field .EGEXECUTE: In this problem Eand cosφare constant over the surface so ()()()22coscos14 N/Ccos600.250 m1.8 Nm /C.EEdAEAφφΦ===°=⋅∫(b) EVALUATE: EΦis independent of the shape of the sheet as long as φand Eare constant at all points on the sheet. (c) EXECUTE: (i) cos. EEEAφΦ=Φis largest for 0 , so cos1 and .EEAφφ=°=Φ=(ii) EΦis smallest for 90 , so cos0 and 0.Eφφ=°=Φ=EVALUATE: EΦis 0 when the surface is parallel to the field so no electric field lines pass through the surface. 22.2. IDENTIFY: The field is uniform and the surface is flat, so use cosEEAφΦ=. SET UP: φis the angle between the normal to the surface and the direction of EG, so 70φ=°. EXECUTE: 2(75.0 N/C)(0.400 m)(0.600 m)cos706.16 Nm /CEΦ==⋅°EVALUATE: If the field were perpendicular to the surface the flux would be 218.0 Nm /C.EEAΦ==⋅The flux in this problem is much less than this because only the component of EGperpendicular to the surface contributes to the flux. 22.3. IDENTIFY: The electric flux through an area is defined as the product of the component of the electric field perpendicular to the area times the area. (a)SET UP: In this case, the electric field is perpendicular to the surface of the sphere, so 2(4)EEAErπΦ==. EXECUTE: Substituting in the numbers gives ()()26521.2510N/C40.150 m3.5310N m /CEπΦ=×=×⋅(b)IDENTIFY: We use the electric field due to a point charge. SET UP: 2014qErπ=PEXECUTE: Solving for qand substituting the numbers gives ()()22660922140.150 m1.2510N/C3.1310C9.0010N m /Cqr Eπ−==×=××⋅PEVALUATE: The flux would be the same no matter how large the circle, since the area is proportional to r2while the electric field is proportional to 1/r2. 22.4. IDENTIFY: Use Eq.(22.3) to calculate the flux for each surface. Use Eq.(22.8) to calculate the total enclosed charge. SET UP: ˆˆ(5.00 N/Cm)(3.00 N/Cm)xz−⋅⋅E =i +kG. The area of each face is 2L, where 0.300 mL=. EXECUTE: 111ˆˆˆ0sSA−⇒ Φ =⋅=n=jE nG. 2222ˆˆˆ(3.00 N C m)(0.300 m)(0.27 (N C) m)SSAzz+⇒ Φ=⋅=⋅=⋅n=kE nG. 22(0.27 (N/C)m)(0.300 m)0.081 (N/C)mΦ ==⋅. 333ˆˆˆ0SSA+⇒ Φ =⋅=n=jE nG. 444ˆˆˆ(0.27 (N/C)m)0 (since 0).SSAzz−⇒ Φ=⋅= −⋅==n=kE nG5525ˆˆˆ(5.00 N/Cm)(0.300 m)(0.45 (N/C)m) .SSAxx+⇒ Φ =⋅= −⋅= −⋅n=iE nG25(0.45 (N/C)m)(0.300 m)(0.135 (N/C)m ).Φ = −⋅= −⋅666ˆˆˆ(0.45 (N/C)m)0 (since 0).SSAxx− ⇒ Φ=⋅= +⋅==n=iE nG22
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