# 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

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
4 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