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# 3 m u 899 j for b vp 100 10 6 c 300 10 6 c kq 1

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Unformatted text preview: e point charge should have the same value Solutions will be posted at www.prep101.com/solutions 13 ©Prep101 Phys1026 April Exam Solutions qpoint = 7.85 x10-7 C b) This point is inside a conductor – the electric field is always zero. The minimum point charge required is zero! ELECTRIC POTENTIAL PRACTICE QUESTIONS Question 1: Solution: The Work done on the electron by the electric field is given by the equation: ( )( W = q Δ V = 1 .6 x10 − 19 3.0x10 3 ) = 4 .8 x10 − 16 J Since the object starts from rest, we can assume that all of this work is converted to 1 kinetic energy of the electron. Thus, we get K = me v 2 = 4.8 x10 −16 J 2 Solving for v we get: v= ( ) 2 4.8 x10 −16 = 3.2 x10 7 m s −31 9.11x10 Question 2: Solution: K= 12 6.7 × 10 −22 (4.8 × 10 5 ) 2 mv 2 v kv mv = ΔU = −qEΔx ⇔ E = = = 1206 = 1.2 −19 m m 2 − 2qΔx − 2 ⋅ 3.2 ⋅ 10 ⋅ (−2m) Question 3: Solution: For A: U = q2V1 = k e q 1q 2 (8.99 ⋅ 10 9 Nm 2 /C 2 )(100 ⋅ 10 −6 C)(−300 ⋅ 10 −6 C) = r12 0.3 m U = -899 J For B: Vp = ⎛ 100 ⋅ 10 −6 C − 300 ⋅ 10 −6 C ⎞ kq 1 kq 2 ⎟ + = (8.99 ⋅ 10 9 Nm 2 /C 2 )⎜ + ⎜ ⎟ r1 r2 1m 0.8 m ⎝ ⎠ Solutions will be posted at www.prep101.com/solutions 14 ©Prep101 Phys1026 April Exam Solutions Vp = -2.47⋅106 V For C: U = -eVP U = (-1.602⋅10-19 C) ⋅ (-2.47⋅106 V) U = 3.96⋅10-13 J Question 4: Solution: For A: r kq 1q 2 (8.99 ⋅ 10 9 )(10 ⋅ 10 −9 )(100 ⋅ 10 −9 ) F= = 8.99x10-6 N = 2 2 r (1) For B: Yes, since the charges have opposite signs. For C: vv v E = E1 + E 2 E= kq 1 r1P 2 + kq 2 r2P 2 = (8.99 ⋅ 10 9 Nm 2 / C 2 ) ⋅ (110 ⋅ 10 −9 C) (0.5m) 2 E = 3956 N/C; where the direction of the electric field is to the right For D: V=− kq 2 (8.99 ⋅ 10 9 Nm 2 / C 2 ) ⋅ (-100 ⋅ 10 −9 C) =− = 899 V r12 (1 m) Solutions will be posted at www.prep101.com/solutions 15 ©Prep101 Phys1026 April Exam Solutions Question 5: Solution: For A: V = kq / r = +900 V q = 900 r / k q= (0.07 m) (900 V) (9 x10 9 Nm 2 /C 2 ) q = +7x10−9 Coulombs For B: Any metallic solid has a constant potential throughout. Therefore, the potential difference between the surface and the center = 0 V For C: For r > 7 cm: V(r) = kq / r (9 ⋅ 10 9 )(7 x10 −9 ) V(21 cm) = = +300 V 0.21 At r = 7 cm: V = 900 V ΔV = V7 – V21 = 900 V – 300 V = +600 V For D: New charge q ′ = Vr (909)(0.07 ) = = 7.07 x10 −9 C 9 k 9x10 ∴Δq = q′ - qinit = (7.07 – 7) x 10−9 Coulombs ∴ΔU = ΔV Δq = (900 V) ⋅ (0.07x10-9 C) = 6.3x10-8 J Question 6: Solution: First, let’s call to the right to be the positive direction For A: Since A is negative, the electric field at P due to A will point toward A (to the left). Since B is positive, the electric field at P due to B will point away from B (to the left as well). −6 9 N ⋅m kQ A 9.00 × 10 c 2 (−2.50 × 10 C ) = −9.00 × 10 4 EA = 2 = 2 rA (0.500m ) 2 N C −6 9 N ⋅m kQB 9.00 × 10 c 2 (5.00 × 10 C ) = 1.80 × 10 5 N C EB = 2 = 2 rB (0.500m ) 2 Solutions will be posted at www.prep1...
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## This document was uploaded on 03/21/2014 for the course PHYSICS 1402 at UWO.

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