Final No Date 2

Final No Date 2

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Unformatted text preview: 01.com/solutions 16 ©Prep101 Phys1026 April Exam Solutions One thing to remember with these types of questions is that we figure out the directions of the individual electric fields and then use the equation to calculate the magnitude only, so that although we got a negative value for EA, we will ignore the negative sign and simply add the two fields using the signs we determined earlier. We determined that both fields are to the left so that with our sign convention they are both negative: E P = − E A − E B = −9.00 × 10 4 N C − 1.80 × 10 5 N C = −2.7 × 10 5 N C 5N Our final answer is − 2.70 × 10 C to toward A For B: With electrostatic potential, we do not ignore the sign we get from our governing equation. Additionally, since the potential is a scalar we can simply add our results from A and B VP = V A + VB VP = VP = VP = kQ A kQB + rA rB k (Q A + QB ) r 2 9.00 × 10 9 N ⋅m c 2 0.500m V P = 4.50 × 10 4 V (−2.50 × 10 −6 C + 5.00 × 10 −6 C ) For C: Here we will use both the force on a charge in an electric field equation and F=ma, where the only force being applied is the one due to the net electric field from part a) r r F = ma r r F = QE ma = QE QE m − 1.60 × 10 −19 C (−2.70 × 10 5 N C ) a= 9.11 × 10 −31 kg a= a = 4.74 × 1016 m s 2 Solutions will be posted at www.prep101.com/solutions 17 ©Prep101 Phys1026 April Exam Solutions Since this is a positive value the final answer is a = 4.74 × 1016 m s 2 toward B. Question 7: Solution: For A: At the origin, we have: V= For B: V= kq − 3kq − 2kq + = a a a kq − 3kq kq − 6 kq − 5kq + = = 4a 2a 4a 4a For C: 2 2 r 2kq 2 ˆ 6kq ˆ 8kq ˆ F= 2 i+ 2 i= 2 i a a a r 8kq 2 F= 2 a For D: This sphere includes the charge -3q, but not the charge q ΦE = ΦE = q enc ε0 − 3q ε0 Question 8: Solution: For A: This is simply the sum of the potentials due to each of the charges where the distance r is given by r= (0.3cm)2 + (0.4cm)2 = 0.5cm V = V20 μC + V−15 μC V= (20μC ) + (− 15μC ) 4πε 0 r 4πε 0 r ⎛ Nm 2 ⎞ 5μC ⎟ V = ⎜ 9 × 10 9 ⎜ C 2 ⎟ 0.5cm ⎝ ⎠ Solutions will be posted at www.prep101.com/solutions 18 ©Prep101 Phys1026 April Exam Solutions ⎛ Nm 2 ⎞ 5 × 10 −6 C V = ⎜ 9 × 10 9 2 ⎟ ⎜ C ⎟ 0.5 × 10 − 2 m ⎝ ⎠ V = 9.0 × 10 6 V For B: Find the electric field at the origin of the x and y axis. rr r E = E 20 μC + E −15 μC r (20μC ) ˆ (− 15μC ) − i ˆ E= i+ 2 4πε 0 r 4πε 0 r 2 r⎛ Nm 2 ⎞ 35μC ˆ i E = ⎜ 9 × 10 9 2 ⎟ ⎜ C ⎟ (0.3cm )2 ⎝ ⎠ r⎛ Nm 2 ⎞ 35 × 10 −6 C ˆ ⎜ 9 × 10 9 2 ⎟ E =⎜ i C ⎟ (0.3 × 10 − 2 )2 ⎝ ⎠ () E = 3.5 × 1010 () Nˆ i C For C: ˆ The force on the positive sphere is in the positive i direction since it is an attractive force due to the charges having opposite signs. r qq ˆ F = k 1 22 r r r⎛ Nm 2 ⎞ (20 μC )(− 15μC ) ˆ i F = ⎜ 9 × 10 9 2 ⎟ ⎜ C⎟ (0.6cm )2 ⎝ ⎠ r ˆ F = 7.5 × 10 4 N i We drop the negative sign since we already argued whic...
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This document was uploaded on 03/21/2014 for the course PHYSICS 1402 at UWO.

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