HW1-phy2049-Spring-2012-solution

00 400 nc 300 nc 2 2 when b is grounded

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Unformatted text preview: 7 m so x = - r = - 0.17m there is no y component of the position by symmetry. 21- 48. (a) Since qA = –2.00 nC, qB = –4.00 nC and qC = +8.00 nC and d= 20cm, Eq. 21- 4 leads to r | q q | | (8.99 ×109 N ⋅ m 2 C2 ) (−2.00 ×10 −9 C)(8.00 ×10 −9 C) | | FAC |= A C 2 = = 3.60 ×10−6 N. 2 4πε 0 d (0.200 m) (b) After making contact with each other, both A and B have a charge of qA + qB ⎛ −2.00 + ( −4.00 ) ⎞ =⎜ ⎟ nC = −3.00 nC. 2 2 ⎝ ⎠ When B is grounded its charge is zero. After making contact with C, which has a charge of +8.00 nC, B acquires a charge of [0 + (8.00 nC)]/2 = 4.00 nC, which charge C has as well. Finally, we have QA = –3.00 nC and QB = QC = +4.00 nC. Therefore, r | q q | | (8.99 ×109 N ⋅ m 2 C2 ) ( −3.00 ×10−9 C)( −4.00 ×10...
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This note was uploaded on 04/02/2013 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.

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