## PHYS208 Fundamentals of Physics II

### Answer for Ch. 33, 25P

a) *U*_{tot} = 1.98 microJ

b) *q*_{max} = 5.56 microC

c) *i*_{max} = 12.6 mA

d) phase angle is -46.9 degrees

Why negative? We found that the cosine of the phase angle should be 0.68;
thus the phase angle may be +/- 46.9 degrees. The sign is chosen depending on
whether the capacitor is charging or discharging. To decide, look at the time derivative of the
charge at *t=0*, which is proportional to the negative of the sine of the phase angle.
For a charging capacitor, this derivative should be positive; thus the choice of -46.9 degrees.

e) phase angle is +46.9 degrees

"http://www.physics.udel.edu/~watson/phys208/exercises/answers/33-25P.html"

Last updated Nov. 25, 1997.

Copyright George Watson, Univ. of Delaware, 1997.