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Two fixed charge A and B of 5uc each are separated by a distance of 6m.C is mid point of line joining A and B.A charge 'Q' of -5uc is shot perpendicular to tge line joining A and B through C with a kinetic energy of 0.06J. the charge 'Q' comes to rest at a point D. the distance CD is

Two fixed charge A and B of 5uc each are separated by a distance of 6m.C is mid point of line joining A and B.A charge 'Q' of -5uc is shot perpendicular to tge line joining A and B through C with a kinetic energy of 0.06J. the charge 'Q' comes to rest at a point D. the distance CD is

Grade:12th pass

1 Answers

Arun
25750 Points
5 years ago
 
 
I suppose the charge  q = - 5 μC starts with a KE of 0.06 J from point C and stops at point d.
Initial total energy of the charge q = KE + PE
  = 0.06 + 9*10⁹ * 5 *10⁻⁶ * (-5 *10⁻⁶)/3  + 9*10⁹ * 5*10⁻⁶*(-5*10⁻⁶)/3 J
  = 0.06 - 0.075 - 0.075  J
  = - 0.09 J
Let ad = bd = x m
Final energy = PE (as KE = 0)
   = - 2 * 9*10⁹ * 5*10⁻⁶ * 5 * 10⁻⁶ /x  J
   = - 0.450/x   J
Hence, using Energy conservation we get:
    0.450/x = 0.09
    x = 5 m
 
cd  = √(x² - 3²) = 4 m
 
 

hope it helps

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