in this problem all that u need to know of electro is that F_qQ=kqQ/R².........

there is a pt charge located at a pt (0,-a) and another pt charge located at a pt (-infinity , 0)

the second pt charge is fired with a velocity of  V₀i........has a mass 'm'....

find the angle by which it is deflected........

involves majority of mechanics more than anything else..........

1st charge is kept fixed........{the one at (0,-a)}....both the chrges are of magnitude 'q'.......

sumone try this one out......

yes......u got it rite.its fired in the x-direction......  with velocity vector V₀i..........

cmon' ppl give it a try naa...

OK...as no one seems to show interest in evn trying this good problem i have decided that i myself shud post the soln....theek hai...a nice problem spoilt congo...

ok the only forces existant here are radial so i net torque=0 and thus dL/dt=0 or i can conserve angular momentum abt the origine....so L_{at (-inf,0)} = L_{after complete defln}

or.......mV₀a = mV_fd.......d is the component of radius vector perp to the velocity...but conserving energy we find out that V_f= V₀ and thus d=a.....

concider the motion at any angle @ (@= alpha)

Accleration vector = KQ²/mR²( -cos@ i + sin@ j)

taking acceleration in y-dirn. we have V_ydV_y/dy = KQ²sin@ /R²----------------------1)

and as said before dL/dt =0 so we can conserve L

mR²d@ /dt = mV₀a-------------------------2)

but d@/dt = d@ /V_ydy.......(mult and divide by dy)-----------3)

so using 2) and 3) we can say dy/R²= V_yd@ /(V₀a)---------------4)

and using 1) and 4) we have

dV_y = KQ²sin @ d@ / (mV₀a)

integrating.....

₀S^{V_osin $} {dV_y} = ₀S^180-$ {KQ²sin @ d@ / (mV₀a)}

(sorry  yaar....my formula edittor isnt working so using notations like S for integration hope u are able to decipher my integration limits)

and thus using the following integrated eqn we can solve for $ ($= theta)....

hope it is useful.....

trying to attach the file......