Halliday-cap 22-vol.3- Rar - P22 036

Transcript

36. (a) Since qA = −2Q and qC = +8Q, Eq. 22-4 leads to

4Q2

|(−2Q)(+8Q)| = .
FAC
= 4π0 d 2 π0 d 2 (b) After making contact with each other, both A and B have a charge of   −2Q + (−4Q) = −3Q . 2 When B is grounded its charge is zero. After making contact with C, which has a charge of +8Q, B acquires a charge of [0 + (−8Q)]/2 = −4Q, which charge C has as well. Finally, we have QA = −3Q and QB = QC = −4Q. Therefore,

3Q2

|(−3Q)(−4Q)| = .
FAC
= 4π0 d 2 π0 d 2 (c) We also obtain

4Q2

|(−4Q)(−4Q)| = .
FBC
= 4π0 d 2 π0 d 2