Homework_03-solutions - toupal(rgt374 Homework 03...

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toupal (rgt374) – Homework 03 – Chiu – (58295) 1 This print-out should have 25 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A dipole (electrically neutral) is placed in an external field. + (a) + (b) + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + (c) (d) For which situation(s) shown above is the net force on the dipole equal to zero? 1. None of these 2. Some other combination 3. (c) only 4. (a) only 5. (a) and (c) 6. (c) and (d) correct 7. (b) and (d) Explanation: The force on a charge in the electric field is given by vector F = q vector E and the torque is defined as vector T = vectorr × vector F Δ vector E = k Δ q r 2 ˆ r vector E = summationdisplay Δ vector E i . Symmetry of the configuration will cause some components of the electric field to be zero. Gauss’ law states Φ S = contintegraldisplay vector E · d vector A = Q ǫ 0 . The electric dipole consists of two equal and opposite charges separated by a distance. In either situation (c) or (d), the electric field is uniform everywhere between the infinite parallel plates. Thus, the electric force on one charge is equal but opposite to that on another so that the net force on the whole dipole is zero. By contrast, electric fields are nonuniform for both (a) and (b). 002 10.0 points Pictured below is a distribution of 6 point charges and their surrounding electric field. + Q - Q + Q - Q - Q + Q Gaussian surface What is the total electric flux through the closed Gaussian Surface shown? 1. Q E 0 2. 2 Q E 0 3. 2 Q E 0 correct 4. 6 Q E 0 5. 0
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toupal (rgt374) – Homework 03 – Chiu – (58295) 2 6. Q E 0 7. 6 Q E 0 Explanation: The total charge within the Gaussian Sur- face is 2 Q , so the total electric flux is φ = 2 Q E 0 . 003 10.0 points A charge moves a distance of 1 . 7 cm in the direction of a uniform electric field having a magnitude of 211 N / C. The electrical potential energy of the charge decreases by 109 . 706 × 10 19 J as it moves. Find the magnitude of the charge on the moving particle. (Hint: The electrical poten- tial energy depends on the distance moved in the direction of the field.) Correct answer: 3 . 05844 × 10 18 C. Explanation: Let : Δ d = 1 . 7 cm , E = 211 N / C and Δ U electric = 109 . 706 × 10 19 J . Δ U electric = q E Δ d q = Δ U electric E Δ d = 109 . 706 × 10 19 J (211 N / C)(0 . 017 m) = 3 . 05844 × 10 18 C . 004 10.0 points Consider two points A and B in a constant electric field vector E as shown. B A 30 0 E What is the magnitude of the potential dif- ference between A and B ? 1. E ℓ 2 2. None of these 3. 0 4. E ℓ 5. 2 E ℓ 6. 2 E ℓ 7. E 8. 2 E ℓ 3 9. E ℓ 2 10. 3 E ℓ 2 correct Explanation: The potential difference between two points A and B is Δ V = integraldisplay B A vector E · dvectors.
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