Phys102 Coordinator: A.A.Naqvi Second Major-132 Sunday, April 20, 2014 Page: 1 Q1. As shown In Figure 1 four particles form a square of side length a. The charges q1 = + Q, q2 = q3 = q, q4 = − 4.00 Q. What is the ratio q/ Q if the net electrostatic force on q1 is zero? F13 Figure 1 A) 1.41 +Q B) 0.707 q C) 2.82 D) 1.00 F12 E) Zero Ans: F14 if net electrostatic force is zero then q -4Q |πΉ13 | = |πΉ14−π¦ | |πΉ13 | = |πΉ14−π¦ | ππ1 π3 ππ1 π4 = cos 45 2 π 2π2 2 |π4 | 4π |π3 | = × πππ 45 ⇒ π = πππ 45 2 2 π = 2ππππ 45 π = 2πππ 45 = 1.41 π Q2. 12 Consider two neutral point particles of mass 5.0 g each. A total of 4.0 × 10 electrons are transferred from one neutral particle to the other particle. How far apart must the two particles be if the magnitude of the electrostatic force between them is equal to the magnitude of the weight of one of the particles on earth surface? A) B) C) D) E) 27 cm 31 cm 33 cm 25 cm 39 cm Ans: ππ = ππ 2 π2 ππ 2 ⇒ π= √ ππ 9 × 109 × (4 × 1012 × 1.6 × 10−19 )2 9 × 16 × 1.6 × 1.6 × 10−2 √ = π 5 × 10−3 × 9.8 5 × 9.8 π = 0.27 m = 27 ππ π= √ Phys102 Coordinator: A.A.Naqvi Second Major-132 Sunday, April 20, 2014 Page: 2 Q3. Three point charges are located on a circular arc as shown in Figure 2. What is the magnitude of net electric field at point P, the center of the arc? +2.00 nC Figure 2 y 4.00 cm x −3.00 nC 30.0o P 30.0o 30ο° 30ο° 4.00 cm +2.00 nC A) B) C) D) E) 3 2.61 × 10 N/C 2.08 × 104 N/C 1.64 × 103 N/C 3.64 × 103 N/C 2.25 × 105 N/C Ans: πΉπππ‘−π¦ = 0 πΉπππ‘−π₯ 2 × π × π+ cos 30° π × π− = − 2 (0.04) (0.04)2 πΉπππ‘−π₯ = π [2 × 10−9 × 2 × πππ 30 − 3 × 10−9 ] (0.04)2 πΉπππ‘ = 9 × 109 × 10−9 [4πππ 30 − 3] (0.04)2 = 9 × 0.464 = 2.61 × 103 N/C (0.04)2 King Fahd University of Petroleum and Minerals Physics Department c-20-n-20-s-0-e-1-fg-1-fo-0 Phys102 Coordinator: A.A.Naqvi Second Major-132 Sunday, April 20, 2014 Page: 3 Q4. An electron enters a region of uniform electric field E = (60 iˆ ) N/C with a velocity vi = (50 iˆ ) km/s. How far does the electron travel in the first 2.0 ×10-9 s time interval after entering the field? A) B) C) D) E) 7.9 ×10-5 m 1.1 ×10-5 m 2.7 ×10-4 m 1.3 ×10-6 m 4.2 ×10-5 m Ans: 1 2 ππ πΈ 1.6 × 10−14 × 60 βπ = π0 π‘ + ππ‘ ⇒ |π| = = = 1.054 × 1013 2 ππ 9.11 × 10−31 |π0 | = 50 km/s = 50 × 103 π/π = 5 × 104 π/π 1 βπ = 5 × 104 × 2 × 10−9 − × (1.054 × 1013 ) × (2 × 10−9 )2 2 1 = 10 × 10−5 − × 1.054 × 4 × 10−5 2 = 7.9 × 10−5 π Q5. A certain electric dipole is placed in a uniform electric field E of magnitude 10 N/C. The magnitude of torque on the dipole plotted as a function of the angle between E and the dipole moment p is shown in Figure 3. How much work is needed by an external agent to turn the electric dipole from 300 to 600 with respect to E field? A) B) C) D) E) +1.83 ×10-1 J −1.83 ×10-1 J +2.66 ×10-1 J −2.66 ×10-1 J +9.20 ×10-2 J Figure 3 Ans: πππ₯π‘ = βπ = ππΈ(πππ ππ − πππ ππ ) ππππ ππππβ ππΈ = 50 × 10−2 πππ₯π‘ = 50 × 10−2 (πππ 30 − πππ 60) πππ₯π‘ = 1.83 × 10−1 J King Fahd University of Petroleum and Minerals Physics Department c-20-n-20-s-0-e-1-fg-1-fo-0 Phys102 Coordinator: A.A.Naqvi Second Major-132 Sunday, April 20, 2014 Page: 4 Q6. A metallic sphere contains a cavity at the center as shown in Figure 4. The outer surface of the sphere is grounded by connecting a conducting wire between it and the earth. A negative point charge Q = −5.4 ×10-9 C is placed inside the cavity of the sphere. What is the net electric flux through the outer surface of the metallic sphere? A) B) C) D) E) Figure 4 0 +6.1×102 N.m2/C −6.1×102 N.m2/C +3.1×102 N.m2/C −3.1×102 N.m2/C Q=0 −Q Ans: ππ π‘βπ π πβπππ ππππ‘ = 0 ππππ‘ = 0 Q7. Consider two infinitely long thin wires carrying uniform linear charge densities λ1 and λ2. The wires are arranged as shown in Figure 5 and λ2 = +5.50 nC/m. If the net electric field at P is zero, determine the magnitude of λ1. Figure 5 P A) 2.75 nC/m B) C) D) E) 1.50 1.75 2.00 0.50 nC/m nC/m nC/m nC/m λ1 λ2 2 cm 2 cm Ans: 2ππ2 2ππ1 πΈπ = 0 ⇒ |πΈ2 | = |πΈ1 | ⇒ | || | π2 π1 |π1 | = |π2 | × π1 π2 2 = 5.5 × 10−9 × ( ) = 2.75 ππΆ/π 4 King Fahd University of Petroleum and Minerals Physics Department c-20-n-20-s-0-e-1-fg-1-fo-0 Phys102 Coordinator: A.A.Naqvi Second Major-132 Sunday, April 20, 2014 Page: 5 Q8. Figure 6 shows two large, parallel, non-conducting sheets, each with fixed uniform charge density: ο³ 1 ο½ + 2.0×10ο6 C/m 2 , ο³ 2 ο½ ο 4.0×10ο6 C/m 2 . The ratio of the magnitude of the electric field at point A to that at point B, (EA/EB), is: Figure 6 A) B) C) D) E) 3.0 0.5 1.0 3.5 1.5 E1 E2 E3 E4 Ans: |π1 | + |π2 | |4.0 + 2.0| × 10−6 π/π2 πΈπ΄ = = |π1 | − |π2 | |2.0 − 4.0| × 10−6 π/π2 πΈπ΅ = 3.0 Q9. Figure 7 shows the cross sectional area of two identical charged solid spheres, 1 and 2, of radius R. The charge is uniformly distributed throughout the volumes of both the spheres. The net electric field is zero at point P, which is located on a line connecting the centers of the spheres, at radial distance R/2 from the center of sphere 1. If the charge on sphere 1 is q1 ο½ 7.8 ο C , determine the magnitude of the charge q 2 on sphere 2. Figure 7 A) 8.8 µC B) 3.2 µC C) 9.3 µC E1 D) 3.5 µC E2 E) 6.8 µC Ans: πππ‘ πππππ‘πππ πππππ ππ‘ π πΈπ = 0 πΈβπ = πΈβ2 + πΈβ1 = 0 |πΈβ2 | = |πΈβ1 | ππ2 2 3π (2) = ππ1 π ππ1 × = π 3 2 2π 2 4π2 π1 9π1 = 2 ⇒ π2 = = 8.8 ππΆ 2 9π 2π 8 King Fahd University of Petroleum and Minerals Physics Department c-20-n-20-s-0-e-1-fg-1-fo-0 Phys102 Coordinator: A.A.Naqvi Second Major-132 Sunday, April 20, 2014 Page: 6 Q10. Two charges, QA = +2 μC and QB = − 6 μC, are located on the x-axis at xA= −1 cm and xB= +2 cm, respectively. Where should a third charge, QC = + 6 μC, be placed on the positive x-axis so that the potential at the origin is equal to zero? A) B) C) D) E) x = 6 cm x = 4 cm x = 7 cm x = 3 cm x = 1 cm + 2 μC O A - 1.0 cm P - 6 μC B + 6 μC D + 2 cm d Ans: ππ = π [ ππ΄ ππ΅ ππ· + + ]=0 0.01 0.02 π ππ· ππ΄ ππ΅ 2 × 10−6 6 × 10−6 2 × 10−6 = − − =− + = π 0.01 0.02 0.01 0.02 0.02 3 ππ· 6 × 10−6 2 × 10−6 3 1 = = ⇒ = π π 0.02 π 0.02 π = 3 × 0.02 = 0.06 π = 6 ππ Q11. The electric potential over a particular region is given by V(x,y)= 9 − 7x − 5y2. Determine the angle between the electric field E at point P and the positive x axis. The coordinates of the point P are x=1.00 m and y=2.00 m. A) B) C) D) E) 70.7 ο° 43.6 ο° 46.6 ο° 83.6 ο° 53.6 ο° Ans: πΈπ₯ (π₯ = 1.00 π) = − ππ£ =7 ππ₯ ππ£ = 10 π¦ = 20 ππ¦ πΈπ¦ 20 π = tan−1 ( ) = tan−1 ( ) = 70.7° πΈπ₯ 7 πΈπ¦ (π¦ = 2.00 π) = − King Fahd University of Petroleum and Minerals Physics Department c-20-n-20-s-0-e-1-fg-1-fo-0 Phys102 Coordinator: A.A.Naqvi Second Major-132 Sunday, April 20, 2014 Page: 7 Q12. A metallic isolated sphere of diameter 3.5 cm, fixed in space, carries +9.0 nC charge. A proton was released from rest at the sphere’s surface. Determine the maximum speed of the proton. (Assume V=0 at infinity) A) B) C) D) E) 9.4×105 1.2×105 3.1×103 4.2×107 3.2×107 m/s m/s m/s m/s m/s Ans: βπΎ = π = −βπ = ππ − ππ ππ’π‘ πΎπ = ππ = 0 πβππ πΎπππ₯ = πΎπ = ππ ⇒ π£πππ₯ = √ πππ π 1 2 ππ π£πππ₯ = 2 π πππ π 2 2 × 9 × 109 × 9 × 10−9 × 1.6 × 10−16 × =√ 1πππ π 1.67 × 10−27 × 0.0175 = 9.4 × 105 π/π Q13. Two metal spheres 1 and 2 with radii r1 = 1.0 cm and r2=2.0 cm carry charges q1= +22 nC, and q2= ο10 nC, respectively. Initially both spheres are far apart. Then the spheres are connected by a thin wire, how much charge is lost by sphere 1 when the electrostatic equilibrium is reached? A) B) C) D) E) +18 −18 +12 −12 +14 nC nC nC nC nC Ans: ππππ‘ = +22 ππ − ππππ = 12 ππΆ After connection π1 = π2 ⇒ ππ1 π(ππππ‘ − π1 ) = π1 π2 π1 ππππ‘ − π1 π1 ππππ‘ = ⇒ π1 π2 = π1 (ππππ‘ − π1 ) ⇒ π1 = π1 π2 π1 + π2 π1 = 1 × 12 × 10−9 = 4.0 × 10−9 C, 3 King Fahd University of Petroleum and Minerals Physics Department ππππ π‘ = (22 − 4.0) × 10−9 = 18 ππΆ c-20-n-20-s-0-e-1-fg-1-fo-0 Phys102 Coordinator: A.A.Naqvi Second Major-132 Sunday, April 20, 2014 Page: 8 Q14. Figure 8 shows a four-capacitor arrangement that is connected to a larger circuit at points A and B. The capacitances are C1 = l0 οF, and C2= C3= C4= 20 οF. The charge on capacitor C1 is 30 οC. What is the magnitude of the potential difference VA− VB? Figure 8 D 30μC A C1+C2 C3+C4 B ⇒ A) B) C) D) E) 5.3 V 3.5 V 2.2 V 1.2 V 7.2 V Ans: ππΆ1 = + π1 30 × 10−6 = = 3 π, ππ2 = ππ1 πΆ1 10 × 10−6 π2 = πΆ2 ππ2 = 20 × 10−6 × 3 = 60 × 10−6 πΆ ππ3+π4 = ππ1+π2 = 60 + 30 = 90 ππΆ ππ3 = ππ3+π4 90 × 10−6 = = 45 ππΆ 2 2 ππ3 = 4.5 × 10−6 /20 × 10−6 = 45 = 2.25 π 20 ππ΄ − ππ΅ = 3 + 2.25 = 5.3 π King Fahd University of Petroleum and Minerals Physics Department c-20-n-20-s-0-e-1-fg-1-fo-0 Phys102 Coordinator: A.A.Naqvi Second Major-132 Sunday, April 20, 2014 Page: 9 Q15. A 200 V battery is connected across 10 identical capacitors which are connected in series. The total energy stored in the combination is 0.0400 J. What is the capacitance of each capacitor? A) B) C) D) E) 20.0 οF 1.00 οF 2.00 οF 6.00 οF 30.0 οF Ans: 1 1πΆ 2 ππ‘ππ‘ = πΆππ π 2 = π π = 10 2 2π 0.04 = πΆ= 1 πΆ 0.04 × 20 × × (200)2 ⇒ πΆ = (200)2 2 10 0.04 × 20 = 20.0 ππΉ (200)2 Q16. A parallel-plate capacitor filled with a material of dielectric constant 3.60, has a capacitance of 1.25 nF. When 5500 V potential difference is applied across the capacitor, the electric field between the plates is 1.60 × 107 V/m. What is the area of the plates? A) B) C) D) E) 0.0135 m2 0.102 m2 0.0194 m2 0.0498 m2 0.387 m2 Ans: πΆ0 = πΆπ , π πβππ π΄ = πΆ0 = π0 π΄ π ππ’π‘ π = π πΈ πΆ0 1 πΆπ π 1 1.25 × 10−9 5500 βπ = β β = × × π0 π0 π πΈ 8.85 × 10−12 3.60 1.60 × 107 π΄ = 0.0135 π2 King Fahd University of Petroleum and Minerals Physics Department c-20-n-20-s-0-e-1-fg-1-fo-0 Phys102 Coordinator: A.A.Naqvi Second Major-132 Sunday, April 20, 2014 Page: 10 Q17. When a dielectric slab is inserted between the plates of one of the two identical capacitors, as shown in Figure 9, which of the following statements describe the situation of C2 correctly? Figure 9 A) B) C) D) E) Potential energy stored in C2 decreases Capacitance of C2 decreases Charge on C2 decreases Potential difference across C2 increases Potential energy stored in C2 increases Ans: ππ = π ⇒ ππ decrease πΆπ π= 1 πΆ π 2 ⇒ decreases 2 π π Q18. The electric potential V(x) versus the position x along a copper wire carrying current is shown in Figure 10. The wire consists of three sections, A, B, and C that differ in radius. Rank the three sections according to the magnitude of the current density, least first. Figure 10 A) C, A, B B) A, B, C C) C, B, A D) B, A, C E) A, C, B Ans: J = ππΈ = π βπ βπ ⇒π½∝ βπ₯ βπ₯ βπ βπ βπ ( ) <( ) <( ) βπ₯ π βπ₯ π΄ βπ₯ π΅ King Fahd University of Petroleum and Minerals Physics Department c-20-n-20-s-0-e-1-fg-1-fo-0 Phys102 Coordinator: A.A.Naqvi Second Major-132 Sunday, April 20, 2014 Page: 11 Q19. The resistance of a wire at 0ο°C is 70.0 β¦. If temperature of the wire increases to 100ο° C, its resistance increases by 50 %. What is resistance of the wire at 120ο°C? (Ignore changes in the dimensions of the wire) A) B) C) D) E) 112 β¦ 211 β¦ 101 β¦ 181 β¦ 150 β¦ Ans: π (100°) = π (0°) × 1.5 = 70 × 1.5 = 105 Ω βR = R 0 + α + β T ⇒ α = (105 − 70) βR = = 0.005 R0 × β T 70 × (100 − 0) π (120°) = π (0°)(1 + α × 120) = 70(1 + 0.005 × 120) = 112 Ω Q20. A copper wire of cross sectional area 2.00 × 10−6 m2 and length 4.00 m has a current of 2.00 A uniformly distributed across its area. How much electrical energy is transferred into thermal energy in 1.00 hour (resistivity of copper = 1.69 × 10-8 ο.m) A) B) C) D) E) 487 J 319 J 727 J 559 J 996 J Ans: π2 × π × π Electrical Energy = π × π‘ = π × π × π‘ = × 3600 π΄ 2 = (2)2 × 1.69 × 10−8 × 4 × 3600 2 × 10−6 = 487 J King Fahd University of Petroleum and Minerals Physics Department c-20-n-20-s-0-e-1-fg-1-fo-0