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HW1-phy2049-Spring-2012-solution

# 2 for particle 1 this means f12 f13 0 or f12 f13

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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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