CHAPTER 23 Problem 57. A proton moving to the right at 3.8!105m/senters a region where a 56 kN/C electric field points to the left. (a) How far will the proton get before its speed reaches zero? (b) Describe its subsequent motion. Solution (a) Choose the x-axis to the right, in the direction of the proton, so that the electric field is negative to the left. If the Coulomb force on the proton is the only important one, the acceleration is ax=e(!E)=m.Equation 2-11, with vox=3.8!105m/sand vx=0,gives a maximum penetration into the field region of x!x0=!vox2=2ax=mvox2=2eE=(1.67!10"27kg)(3.8!105m/s)22(1.6!10"19C)(56!103N/C)=1.35 cm.(b) The proton then moves to the left, with the same constant acceleration in the field region, until it exits with the initial velocity reversed. Problem 65. A dipole with dipole moment 1.5 nC!mis oriented at 30°to a 4.0-MN/C electric field. (a) What is the magnitude of the torque on the dipole? (b) How much work is required to rotate the dipole until it’s antiparallel to the field? Solution (a) The torque on an electric dipole in an external electric field is given by Equation 23-11; !=p"E=pEsin#=(1.5 nC!m)(4.0 MN/C) sin30° =3.0 mN!m.(b) The work done against just the electric force is equal to the change in the dipole’s potential energy (Equation 23-12); W=!U=("p#E)f"("p#E)i=pE(cos 30°"cos180°)=(1.5 nC#m)$(4.0 MN/C)(1.866)=11.2 mJ.Chapter 24 Problem 17. The electric field at the surface of a uniformly charged sphere of radius 5.0 cm is 90 kN/C.What would be the field strength 10 cm from the surface? Solution The electric field due to a uniformly charged sphere is like the field of a point charge for points outside the sphere, i.e., E(r)ª1=r2for r!R.Thus, at 10 cm from the surface, r=15 cmand E(15 cm)=(5=15)2E(5 cm)=(90 kN/C)=9=10 kN/C.Problem 18. A solid sphere 25 cm in radius carries 14 μC,distributed uniformly throughout its volume. Find the electric field strength (a) 15 cm, (b) 25 cm, and (c) 50 cm from the sphere’s center.
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