P26_050.PDF

50. (a) We calculate the charged surface area of the cylindrical volume as follows:

A =2 πrh + πr 2 =2 π (0 . 20m)(0 . 10m) + π (0 . 20m) 2 =0 . 25 m 2

0 . 50 µ C on the exterior surface, and consequently (according to the

assumptions in the problem) that same charge q is induced in the interior of the ﬂuid.

(b) By Eq. 26-21, the energy stored is

U = q 2

2 C = 5 . 0 × 10 − 7 C 2

2(35

10 − 12 F) =3 . 6

10 − 3 J .

probably not cause a spark; however, there is not enough of a safety factor to be sure.

where we note from the ﬁgure that although the bottom is charged, the top is not. Therefore,

thechargeis q = σA =

−