As shown in the figure below, a uniform electric field has magnitude E = 290 N/C and is directed to the right. A particle with charge +3.6 nC moves along the straight line from a to b.
(a) What is the electric force that acts on the particle?
b) What is the work done on the particle by the electric field?
(c) What is the potential difference Va - Vb between points a and b?
2006-10-11 17:14:14 · 2 answers · asked by p_rob22 1 in Science & Mathematics ➔ Physics
a) by Coulomb's Law,
F = kqQ/r^2, where k = 1/4*pi*sigma (sigma being the permitivity constant of air).
b) You can figure this out from mechanics, work = F * d, in this case you should integrate the force you figured out from Coulomb's Law from a to b.
c) the negative gradient of V is the electric field of E, which means the potential at any point is given by
V= kQ/r. So, just find V for r=a & r=b then take the difference.
2006-10-11 17:25:26 · answer #1 · answered by Kerintok 2 · 1⤊ 0⤋
(a) Lightning
(b) Pothole repair
(c) 42
OK OK, you got me. I have no idea. But you have successfully helped me decide whether I should take a Physics class for fun LOL! I mean, I love Hawking and Sagan and Schroedinger's cat and Turing Tests and all, but wow. That just sounds like engineering!
Good luck, sorry I can't help.
2006-10-12 00:22:55 · answer #2 · answered by tagi_65 5 · 0⤊ 0⤋