1) A dipole with dipole moment 1.5 nC*m is oriented at 30 degrees to an electric
ID: 2012882 • Letter: 1
Question
1) A dipole with dipole moment 1.5 nC*m is oriented at 30 degrees to an electric field, 4.0*10^6 N/C. (a) What is the torque on the dipole? (b) How much work is required to rotate the dipole until it is parallel to the field (hint: think about work-energy relationship)?(2) Two parallel-plate capacitors with different plate separation but the same capacitance are connected in series to a battery. Both capacitors are filled with air. The quantity that is NOT the same for both capacitors when they are fully charged is (please explain):
(A) potential difference
(B) stored energy
(C) electric field between the plates
(D) charge on the positive plate
(E) dielectric constant
(3) 20 loops of wire with a radius of 7.5 cm rotates about a diameter in a uniform magnetic field of 1.6 T. To produce a maximum emf of 20.0 V, it should rotate at:
(A) 0
(B) 2.7 rad/s
(C) 5.6 rad/s
(D) 35 rad/s
(E) 71 rad/s
Explanation / Answer
Electric Dipole moment P = 1.5 nC m = 1.5*10^-9 C m Electric field E = 4*10^6 N/C Angle between dipolemoment P and Electric field E, = 300 (a) Torque acting on the Dipole, = P X E = PE sin = (1.5*10^-9 C m)(4*10^6 N/C) sin30 = 3*10^-3 N m (b) Work done W = PE cos = (1.5*10^-9 C m)(4*10^6 N/C) cos30 = 5.2*10^-3 J Please post the other questions seperately.Our rules not allowed us to answer more than one question in one post.Thank you so much. (a) Torque acting on the Dipole, = P X E = PE sin = (1.5*10^-9 C m)(4*10^6 N/C) sin30 = 3*10^-3 N m (b) Work done W = PE cos = (1.5*10^-9 C m)(4*10^6 N/C) cos30 = 5.2*10^-3 J Please post the other questions seperately.Our rules not allowed us to answer more than one question in one post.Thank you so much.Related Questions
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