P22_036 - 36. (a) Since qA = −2Q and qC = +8Q, Eq. 22-4...

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Unformatted text preview: 36. (a) Since qA = −2Q and qC = +8Q, Eq. 22-4 leads to FAC = 4Q2 |(−2Q)(+8Q)| = . 4π 0 d 2 π 0 d2 (b) After making contact with each other, both A and B have a charge of −2Q + (−4Q) 2 = −3Q . 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, FAC = 3Q2 |(−3Q)(−4Q)| = . 4π 0 d 2 π 0 d2 FBC = 4 Q2 |(−4Q)(−4Q)| = . 4π 0 d 2 π 0 d2 (c) We also obtain ...
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This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.

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