# UT PHY 317L - HW 1-solutions (6 pages)

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## HW 1-solutions

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## HW 1-solutions

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Pages:
6
School:
The University of Texas at Austin
Course:
Phy 317l - General Physics II
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mughawish am72963 HW 1 coker 56000 This print out should have 12 questions Multiple choice questions may continue on the next column or page find all choices before answering This assignment covers the material from Chapter 20 in the text Coulomb force electric field direct integration over charge distributions the dipole properties of the E field and electrical properties of materials 1 q a Q 2 2 d 2 a 1 d 3 d q a Q2 4 d a q a Q q 6 d 2a Q 5 d 001 10 0 points A particle of mass 17 g and charge 18 C is released from rest when it is 21 cm from a second particle of charge 17 C Determine the magnitude of the initial acceleration of the 17 g particle Correct answer 3668 37 m s2 Explanation Let m 17 g q 18 C 1 8 10 5 C d 21 cm 0 21 m Q 17 C 1 7 10 5 C ke 8 9875 109 N m2 C2 7 d 0 a 2 a 9 d correct 2 Explanation We have the force on charge q on the x axis due to charge Q on the y axis 8 d F and The force exerted on the particle is q Q ma d2 q Q a ke m d2 8 9875 109 N m2 C2 1 8 10 5 C 1 7 10 5 C 0 017 kg 0 21 m2 F ke 3668 37 m s2 002 10 0 points This question will help you review both basic vector concepts and basic calculus concepts Charge Q is on the y axis a distance a from the origin and charge q is on the x axis a distance d from the origin What is the value of d for which the x component of the force on q is the greatest 1 qQ r 4 0 r 2 p where r a2 d2 so the x component of the force on q is 1 4 0 1 4 0 1 4 0 Fx qQ cos r2 qQ d 2 2 a d a 2 d2 qQd 2 a d2 3 2 For maximum x component of the force Fx 0 is required so d Fx qQ a 2 2 d2 0 d 4 0 a2 d2 5 2 a 2 2 d2 0 a d 2 003 10 0 points A tiny charged sphere behaves like a point mughawish am72963 HW 1 coker 56000 charge when you are well away from the sphere Two small spheres carry electric charges of equal magnitudes There are equally spaced points a b and c which lie along the same line a c b What is the direction of the net electric field at each point due to these charges 1 a b c correct 2 a b a b c a b c c a 2 b c Explanation Since the field originates from positive charges and terminates on the negative charges a b c 004 part 1 of 2 10 0 points Two point charges 2 6 pC and 2 6 pC are separated by 7 m What is the dipole moment of this pair of charges Correct answer 1 82 10 17 C m Explanation 3 4 Let q1 2 6 pC 2 6 10 12 C q2 2 6 pC 2 6 10 12 C L 7 m 7 10 6 m so The dipole moment is p q L p q L 2 6 10 12 C 7 10 6 m 1 82 10 17 C m 5 a b c a b c 6 005 part 2 of 2 10 0 points Sketch the pair of charges including the direction of the dipole moment 1 7 a b q c p q p 2 8 q a b c 3 9 a 10 and b c q p q p 4 correct q q q mughawish am72963 HW 1 coker 56000 Explanation 006 10 0 points A dipole electrically neutral is placed in an external field a 3 some component of the electric field to be zero Gauss law states S b I dA Q E 0 c d For which situation s shown above is the net force on the dipole equal to zero 1 b and d 2 b c and d 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 and parallel everywhere 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 007 10 0 points Remember that dE kdqb r r 2 You have to do the integral A uniformly charged insulating rod of length 18 5 cm is bent into the shape of a semicircle as in the figure 3 c and d correct 6 45 C 18 5 4 a b and c 5 a only O 6 None of these 7 c only If the rod has a total charge of 6 45 C find the horizontal component of the electric field at O the center of the semicircle Define right as positive The value of the Coulomb constant is 8 98755 109 N m2 C2 8 a and d 9 Another combination 10 a and c Explanation The force on a charge in the electric field is qE and the torque is defined as given by F T r F E k q r r2 and E X i E Symmetry of the configuration will cause Correct answer 1 06423 107 N C Explanation Let q 6 45 C 6 45 10 6 C L 18 5 cm 0 185 m and k 8 98755 109 N m2 C2 mughawish am72963 HW 1 coker 56000 If the linear charge density is given by 0 x0 x where 0 is a positive constant find the electric field at the origin denotes the unit vector in the positive x direction y b x Ex dE sin ke Z dq sin r2 where dq dx r d so that ke Ex r Z 3 2 where cos d 2 ke sin r 3 2 2 2 k 0 2 x20 k 0 correct 2 2 x0 k 0 3 2 x0 k 20 4 2 x0 k 0 5 x0 k 0 6 x0 Explanation This is a continuous distribution of charge as opposed to point charges We must divide the distribution into small elements and integrate Using Coulomb s law the electric field created by each small element with charge dq is k dq dE 2 x where 1 Call the length of the rod L and its charge q Due to symmetry Z Ey dEy 0 and Z 4 ke r q L and r and L 2 ke q L2 2 8 98755 109 N m2 C2 0 185 m 2 6 45 10 6 C Ex 1 06423 107 N C Since the rod has negative charge a positive …

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