{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

hw_ch19 - Instructor Tomoyuki Nakayama Tuesday PHY 2005...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
Instructor: Tomoyuki Nakayama Tuesday, May 18, 2010 PHY 2005 Applied Physics 2 - Summer C 2010 Solutions for Suggested Homework Problems (Chapter 19) ________________________________________________________________________________ 1. Each electron has a charge of –e, where e = 1.6×10 -19 C. Thus in 1-C charge, there are 1 C/ (1.6×10 -19 C/electron) = 6.25×10 18 electrons. 7. A proton has a charge +e and an electron has a charge –e . They are unlike charges, thus the force on the electron is toward the proton. The magnitude of the force is given by Coulomb’s law. Coulomb’s law yields F = ke 2 /r 2 = 8.20×10 -8 N 9. (a) Applying Coulomb’s law, we obtain F = k|q 1 ||q 2 |/r 2 = 4.32×10 -5 N, where q 1 = 0.3 μ C and q 2 = -0.4 μ C. The force is attractive because they have opposite signs. (b) After the two identical spheres are touched together, the charge distributes itself so that each sphere has an equal amount of charge. Each sphere has a charge of q = (q 1 + q 2 )/2 = -0.05 μ C. Coulomb’s law yields F = k |q | 2 /r 2 = 9.00×10 -7 N. 11. (a) Electric force is proportional to 1/r 2 . We use the ratio of the force at r 1 = 4 m to the force at r 2 = 3 m to obtain F(r 2 )/F(r 1 ) = (1/r 2 2 )/(1/r 1 2 ) F(r 2 ) = (r 2 2 / r 1 2 )F(r 1 ) = 4.44×10 -8 N .
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}