Chapter 17 Answers

Answers to Even Numbered Problems

All answers have been checked against the answer key, and should be presumed to be correct. You should ask for help in the recitations if you are unable to obtain these results.

2.   -150 eV = -2.40 x 10-17 J

10.   (a) 1.62 x 107 m/s (b) 3.50 x 107 m/s

20.   2.33 x 107m/s

22.   (a) 27.2 V, (b) 13.6 eV, (c) -13.6 eV, (d) 13.6 eV

38.   0.24 m2

40.   C = 4,315 pF, A = 0.254 m2

46.   The energy doubles.

Answers to Questions

1. The work is the change in potential energy, so no work is done to move a test charge from one point to another at the same electric potential. This does not imply that not force is exerted. No force if exerted if the test charge moves along an equipotential surface at all times. However, if it moves across equipotential surfaces and back to the same potential, then a force must be exerted, but the total work done cancels out in the end. This is analogous to walking up a hill and back to the bottom. Work is done going up the hill, but lost on the way back down, so the net work is zero even though force was exerted.

     
2. If two equipotential lines (which really are surfaces in 3 dimensions) cross, the the potentials must be the same where they cross, which means the potentials must be the same for both equipotential surfaces. Therefore, two equipotential surfaces can cross only if the potentials are the same. For example, consider charges +1C at (x,y) = (1,1) and (-1, -1), and charges -1C at (x,y) = (1,-1) and (-1,1). Then the x and y axes are equipotentials at V = 0. See the figure to the right. Example for Question 2  
3. Equipotentials for Question 3   
The equipotentials are perpendicular to the field lines in Figure 16-29(b). They look like contour lines around two hills of equal height centered at the positive charges, with a saddle point between the two hills.  

4. The electric field is zero at a point half way between two equal positive charges, since the net force on a test charge there is zero. The electric potential is zero at infinity, by convention. It is not zero between the two charges, because it takes work to bring a positive test charge from infinity to a point between the charges.

5. The kinetic energy gained is proportional to the voltage, and the speed gained is proportional to the square root of the kinetic energy, so the speed gained is proportional to the square root of the voltage. The final speed would be twice as great if the voltage is increased to 400V.

6. A negative charge will move toward a region of higher potential, and a positive charge will move toward a region of negative potential. The potential energy of the charge decreases in either case.

7. (a) Ane electric potential is the amount of potential energy per unit charge of a small test charge in a field. The electric field is the force per unit charge on the small positive test charge. The potential is a scalar, and the field is a vector. (b) The electric potential is the is the potential energy per unit charge of a small positive test charge in and electric field.

8. The point where the potential is zero can be chosen arbitrarily, since only potential differences are physically relevant, representing the work it takes to move a positive test charge from one point to another. This is not connected to the electric field being zero in any way. If the electric field is zero in a region, all you can say is that the electric potential is constant there. For example, the electric potential is zero at all points along the x and y axes in the figure for question 2, but the field is zero only at the origin or infinity.

9. The electric field is zero in a region of constant potential, since moving a charge about requires no work, and therefore no force.

10. The earth's gravitational field varies as 1/r2, and the gravitational potential varies as 1/r.

11. Yes, this will be the case for a negative charge, since a region of high potential repels positive test charges, but attracts negative charges.

12. (a) V would be decreased by 10V everywhere else. (b) The electric field E would not be affected since it depends only on potential differences.

13. If the charge were of different magnitude, there would be a net charge on the capacitor which would create a large electic field outside the capacitor of magnitude kQ/r2. This would require more energy than an electrically neutral capacitor, so the plates acquire only enough charge to become electrically neutral. This does not depend on the size or shape of the plates.

14. The capacitance does not depend on the amount of charge or the voltage. It is a property of the capacitor only.

15. (a) If Q doesn't change, then the energy is U = Q2/2C. Inserting a dielectric increases C, so it decreases the stored energy. (b) If V doesn't change, then the energy is U = CV2/2. Increasing C by inserting a dielectric then increases the stored energy.

Physics 222 Department of Physics University of Tennessee