Coulomb1 - Coulombs Law. The magnitude of the force is...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
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
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Coulombs Law. The magnitude of the force is given by Coulombs 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....
View Full Document

This document was uploaded on 11/01/2011 for the course PHY PHY2049 at Broward College.

Ask a homework question - tutors are online