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Unformatted text preview: homework 03 – KIM, JI – Due: Feb 3 2008, 4:00 am 1 Question 1, chap 1, sect 1. part 1 of 1 10 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. Another combination 2. ( c ) only 3. ( a ) and ( c ) 4. ( a ) only 5. ( b ) and ( d ) 6. ( c ) and ( d ) correct 7. None of these Explanation: Basic Concepts: Field patterns of point charge and parallel plates of infinite extent. 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 component of the electric field to be zero. Gauss’ law states Φ S = contintegraldisplay vector E · d vector A = Q ǫ . Solutions: 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 parallel infinite 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 situations both ( a ) and ( b ). Question 2, chap 24, sect 2. part 1 of 1 10 points Consider the following 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. Φ = 2 Q E correct 2. Φ = 0 3. Φ = 2 Q E 4. Φ = Q E 5. Φ = 6 Q E 6. Φ = Q E homework 03 – KIM, JI – Due: Feb 3 2008, 4:00 am 2 7. Φ = 6 Q E Explanation: The total charge within the Gaussian sur face is 2 Q , so the total electric flux is Φ = 2 Q E . . Question 3, chap 23, sect 2. part 1 of 1 10 points A uniformly charged insulating rod of length 8 . 1 cm is bent into the shape of a semicircle as in the figure. The value of the Coulomb constant is 8 . 98755 × 10 9 N m 2 / C 2 . 8 . 1 c m 8 . 01 μ C O If the rod has a total charge of 8 . 01 μ C, find the horizontal component of the electric field at O , the center of the semicircle. Define right as positive. Correct answer: 6 . 8942 × 10 7 N / C (toler ance ± 1 %). Explanation: Let : L = 8 . 1 cm = 0 . 081 m , q = 8 . 01 μ C = 8 . 01 × 10 − 6 C , and k e = 8 . 98755 × 10 9 N m 2 / C 2 . Call the length of the rod L and its charge q . Due to symmetry E y = integraldisplay dE y = 0 and E x = integraldisplay dE cos( π θ ) = k e integraldisplay dq cos θ r 2 , where dq = λdx = λr dθ , so that b y x θ E x = k e λ r integraldisplay 3 π/ 2 π/ 2 cos θ dθ = k e λ r (sin θ ) vextendsingle vextendsingle vextendsingle vextendsingle 3 π/ 2 π/ 2 = 2 k e λ r , where λ = q L and r = L π . Therefore, E x = 2 k e q π L 2 = 2 (8 . 98755 × 10 9 N m 2 / C 2 ) (0 . 081 m) 2 × ( 8 . 01 × 10 − 6 C) π = 6 . 8942...
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This note was uploaded on 05/05/2008 for the course PHY 303L taught by Professor Turner during the Spring '08 term at University of Texas.
 Spring '08
 Turner
 Physics, Work

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