P22_009 - F q = 0 and solve for q : q = 4 qx 2 L 2 = 4 9 q...

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9. (a) If the system of three charges is to be in equilibrium, the force on each charge must be zero. Let the third charge be q 0 . It must lie between the other two or else the forces acting on it due to the other charges would be in the same direction and q 0 could not be in equilibrium. Suppose q 0 is a distance x from q , as shown on the diagram below. The force acting on q 0 is then given by F 0 = 1 4 πε 0 µ qq 0 x 2 4 qq 0 ( L x ) 2 where the positive direction is rightward. We require F 0 = 0 and solve for x . Canceling common factors yields 1 /x 2 =4 / ( L x ) 2 and taking the square root yields 1 /x =2 / ( L x ). The solution is x = L/ 3. •• q 4 q q 0 ←− x −→←−− L x −−→ The force on q is F q = 1 4 πε 0 µ qq 0 x 2 + 4 q 2 L 2 . The signs are chosen so that a negative force value would cause q to move leftward. We require F
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Unformatted text preview: F q = 0 and solve for q : q = 4 qx 2 L 2 = 4 9 q where x = L/ 3 is used. We now examine the force on 4 q : F 4 q = 1 4 4 q 2 L 2 + 4 qq ( L x ) 2 = 1 4 4 q 2 L 2 + 4( 4 / 9) q 2 (4 / 9) L 2 = 1 4 4 q 2 L 2 4 q 2 L 2 which we see is zero. Thus, with q = (4 / 9) q and x = L/ 3, all three charges are in equilibrium. (b) If q moves toward q the force of attraction exerted by q is greater in magnitude than the force of attraction exerted by 4 q . This causes q to continue to move toward q and away from its initial position. The equilibrium is unstable....
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This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.

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