Two fixed charges A and B of 5 μ C each are separ

Question

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 · Accepted Answer

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)

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