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Coulomb1 - Coulomb’s Law The magnitude of the force is...

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Coulomb’s Law Example (1-dimension) Consider 2 point charges, q 1 and q 2 , separated by a distance x . Let: 1 2 3 0 ( is some positive number) (both charges are negative) 1 m q q q q q q x = - = - = Convention: ! F 12 denotes the force acting on particle 1 from the presence of particle 2 Since the force is repulsive (same-sign electric charges), ! F 12 = ! F 12 ˆ i (points in negative x direction. Some textbooks denote this by ˆ x ). By Newton’s 3 rd Law, that for every action (force) there is an equal and opposite reaction, the force on particle 2 from particle 1 is: ! F 21 = ! ! F 12 = F 12 ˆ i (points in positive x direction). Again, use this line of reasoning to determine the direction of the force and not the sign in
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Unformatted text preview: Coulomb’s Law. The magnitude of the force is given by Coulomb’s Law: 1 2 2 12 12 2 F | | K | q || q | 3 Kq x = F = = Since the force is non-zero and repulsive, the charges will accelerate in the directions specified by the forces. Can we add a third charge to counteract this force and leave all charge stationary? Yes! Place a third positive charge between charges 1 and 2. Note that the superposition principle holds: adding a third charge does not affect the force between charges 1 and 2. Let particle 1 reside at x = 0, particle 2 at x = 1 m, and particle 3 at x = r , where 0 < r < 1....
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