Final No Date 2

Choose as the gaussian surface a cylinder of length l

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Unformatted text preview: maximum flux, Φ E = EA . Since A = π r 2 , we can write down the magnitude of the field as ΦE 5 .2 x10 5 Nm 2 C = 2 A ⎛ 0 .4 m ⎞ π⎜ ⎟ ⎝2⎠ E = 4 .1 x10 6 N C E= Question 2: Solution: For A: With δ very small, all points on the hemisphere are nearly at distance R from the charge, kQ so the field everywhere on the curved surface is e 2 radially outwards. Then the flux is R this field strength time the area of the sphere: Solutions will be posted at 11 ©Prep101 Phys1026 April Exam Solutions ( ) Q 1 ⎛ k Q ⎞⎛ 1 ⎞ Φ CURVED = ⎜ e 2 ⎟ ⎜ ⎟ 4π R 2 = Q (2π ) = 4πε o 2ε o ⎝ R ⎠⎝ 2 ⎠ For B: The closed surface encloses zero charge, therefore Gauss’ Law gives: Φ CURVED + Φ FLAT = 0 Φ FLAT = − Φ CURVED = − Q 2ε o Question 3: Solution: If ρ is positive, the field must everywhere be radially outward. Choose as the Gaussian surface a cylinder of length L and radius r, contained inside the charged rod. Its volume is πr2L and it encloses charge ρ πr2L. The circular end caps have no electric flux through them; there Φ E = ∫ E ⋅ dA = 0 . The curved surface has Φ E = ∫ E ⋅ dA = ∫ EdA cos θ , and E must be the same strength everywhere over the curved surface. Then Φ E = ∫ E ⋅ dA = q εo ∫ becomes Φ E = E dA = ρπ r 2 L εo Noting that 2πrL is the lateral surface area of the cylinder, E (2π r )L = E= ρπ r 2 L εo ρr 2ε o radially away from the axis. Question 4: Solution: For a large insulating sheet E= σ 2ε o = 2π k e σ So Solutions will be posted at 12 ©Prep101 ( Phys1026 April Exam Solutions )( ) E = 2π 8 .99 x10 9 Nm 2 C 2 8 .99 x10 −6 C m 2 = 5 .09 x10 5 N C E is perpendicular to the sheet. Answer 4.5 C. The field just above the surface is given by E = σ/2ε, where σ = q/A. So we must find which increases σ the most. A. Increasing the charge by 4 μC increases the field by 1.4 B. This will have no effect – the spheres have the same charge. C. Add +6 μC into the cavity, there must be -6 μC on the inner surface of the cavity to prevent an electric field from appearing in the conductor. By conservation of charge, there must now be 16 μC on the surface, increasing the field by 1.6 D. This will eliminate the electric field, by the same logic as C. E. The area is πr2. Reducing the radius to 0.9 reduces the area to 0.81 of the old value, increasing σ and the field by 1/0.81 = 1.23 C is the biggest increase. Answer 4.6 A, C, D, B A is the greatest, right next to a large charge. B is zero, inside a conductor. C is not zero, as insulators allow some electric field. Whatever C is, it must be bigger than D which is much further away, and D is still bigger than zero because all the charges present are the same sign. Answer 4.7 a) This is outside the sphere, so by Gauss’ law we can treat it as a point charge located at its center. The total charge is qsphere= σA = σπr2 = 1 x 10-6 x π x (0.5)2 = 7.8 5 x 10-7 C To cancel this out halfway between the two, th...
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This document was uploaded on 03/21/2014 for the course PHYSICS 1402 at UWO.

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