Question

The two permanent charge A and B of 5 uC are placed at a distance of 6m. The middle point of the line connecting C, A and B is. A charge of 5 uC is towards the side of the line connecting the 'Q' to the line A and B towards 0. 06 J is left with the kinetic energy. is . The charge 'Q' will be in a break at point A. The distance of the CD will be
(a) 3m
(b) ✓3m

(c) 3 √3m
(d) 4m
solve this question please​

Answer 1

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

so d is your answer.

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