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Unformatted text preview: 4 μC, Similar to lecture except that the two charges given here are opposite charge. We will need a third charge at position x=  r (to the left of the origin). Equilibrium ⇒ ∑ i Fji = 0 Sum of all forces acting on particle j equals 0. 2 For particle 1 this means: F12 + F13 = 0
or F12 = −F13 So: q q 
q q 
+ K 1 2 2 ˆ − K 1 2 3 ˆ = 0 r = separation between 1 and 3
i
i
L
r ⇒ q3 = ( r / L )  q2 
2 Equilibrium of particle 2 implies: F21 + F23 = 0 or F21 = −F23 PHY2049 Spring 2012 −K  q2 q1  ˆ
 q2 q3  ˆ
i+ K
i =0
2
2
L
(L + r) 1 + r = separation between 2 and 3, r>0 HW1 ⇒ q3 = (1 + r / L )  q1 
2 Setting the two equations we obtained for q3 equal yields: ( r / L) 2  q2 = (1 + r / L )  q1 
2  q2  /  q1 = ( L / r + 1) 2 L / r + 1 =  q2  /  q1 
r= L
 q2  /  q1  − 1 = 0.1...
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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.
 Spring '08
 Any
 Physics, Force, Work

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