IIT JEE , AIEEE Forum - Physics , Electricity

vali116

A partcle of charge 1C is at rest.A body of mass 200gm and charge 1C is approching this partcle with velocity 1.5INTO10 POWER 5m/s.This body will

(A)cross the charge (B)carry the charge with it

(C)stick to the charge (D)stop at a distance of 4m from the charge

umang

I think it would carry the charge with it .

Nithy

i think, it would stop at a distance 2m from the other charge.

point where K.E cancels repulsive force.

equating, to find r...

correct?

elessar_iitkgp

Let the mass of the incoming particle be m(=200g = 0.2kg), and its velocity be v₀(=1.5 x 10⁵ m/s). Let its charge be q (= 1C)
Also let the charge of the particle at rest be Q( = 1C)
Note that the problem doesn't mention the mass of the particle at rest. So most probably the particle at rest doesn't move, ie, is fixed. Assuming it is so, we conserve energy for the two charge system
(1/4

)(Qq/x) + (1/2)mv² = (1/2)mv₀²
where x is the instantaneous distance between the two charges, and v is the velocity of the incoming charge when its velocity is v.
v² = v₀² - (1/4

)(2Qq/mx)
As v²

0
x

(1/4

)(2Qq/mv₀²)
x

4 m

Hence D

I had earlier taken velocity to be 5 instead of 1.5

nikhiljee2007

ANS (D)

let us assume it stops at a distance : x

0.5*0.2*2.25*(10 raised to 10) = change in KE
(k*1*1/x)-0 = change in PE
[assuming the particle comes from infinity]

EQUATE:-
9*(10 raised to 9)/x = 2.25(10 raised to 9)
x= 9/2.25 = 4m

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