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

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

Grade:12th pass

2 Answers

Arun
25750 Points
6 years ago
Dear student
 
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
 
 
 
Regards
Arun (askIITians forum expert)
Kushagra Madhukar
askIITians Faculty 628 Points
3 years ago
Hello Student
As per the question both have the same charges q = 5 uC
Q = – 5uC
Let the distance of D from the center of B and A be r
according to pythagoras theorem, CD = √(x² - 3²) 
 
Since only electric force acts on the bod, which is conservative.
Hence, Total mechanical energy will be conserved between C and D
 
KC + UC = KD + UD
0.06 + kqAQ/3 + kqBQ/3 = 0 + kqAQ/r + kqBQ/r
0.06 + 2x(9 x 109)x(5x10-6)x(- 5x10-6)/3  = 2x(9 x 109)x(5x10-6)x(- 5x10-6)/r
 
on solving, we get, r = 5m
Therefore, CD =  √(5² - 3²) = 4m
 
Hope it helps,
Regards
Kushagra
 

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