Question

Two fixed charges A and B of $5\mu C$ 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 $-5\mu C$ 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?

Answer 1

I suppose the charge $q=-5\mu C$ starts with a KE of $0.06J$ from point C and point D.

Initial total energy $q=KE+PE$

= $0.06+\frac{9\times {10}^9\times 5\times {10}^{-6}\times (-5)\times {10}^{-6}}{3}+\frac{9\times {10}^9\times 5\times {10}^{-6}\times (-5)\times {10}^{-6}}{3}$

= $0.06-0.075-0.075J$

= $-0.09J$

$ad=bd=xm$

Final energy = PE (as KE = 0)

= - $\frac{2\times 9\times {10}^9\times 5\times {10}^{-6}\times 5\times {10}^{-6}}{x}$

=$-0.450/xJ$

Energy conservation :

$0\frac{.450}{x}=0.09$

$x=5m$

cd = $\sqrt{x^2-3^2}=4m$