IIT JEE , AIEEE Forum - Physics , Electricity

nikhil_ramaswamy

An electron is projected with an initial velocity 3.24* 10^5 m/s directed towards a proton, which is initially at rest. If the electron is initially at a great distance from the proton, at what distance from the proton is its speed instantaneously equal to twice its initial value?

smartgoels

apply conservation of energy

nikhil_ramaswamy

Ya but realise that both the proton and the electron start moving. The motion as such is not really simple as the accelerations are different and the acceleration at any instant depends on the distance between the two. I found it tough to correlate all this. By the way the solution given in the study material in which I came across this problem conviniently ignored the K.E. of the proton and arrived at an answer. I am not sure if that was meant to be an approximation, neither do I know if the approximation is valid if done.

malay

you could apply law of conservation of momentum here

malay

nikhil
I have something for you.

first notation, since I can't use usual notations due to inability

a: mass of electron
b:mass of proton
e:charge of proton
r:the required distance
v:initial velocity of electron
w:final velocity of proton

the question is one dimensional interaction between proton and electron. The system consists of electeon and proton, and they attract each other through conservative electric field(magnetic effects are neglected).

Applying law of consevation of momentum
av=a*2v-b*w
(if electron is moving initiallly in positive x-direction, then finally it will be moving in +x and proton in -x due to mutual attraction)

Aplplying law of conservation of energy
a*v*v/2=a*2v*2v/2+b*w*w/2-eV(r)

Using, these two equations, you can find r

if you find w from first equation and substitute in second equation, you observe that kinetic energy of proton involves square of a/b, which is very small and hence can be neglected.

edison

Since the electron and proton are oppositely charged

proton can be assumed to be at rest as mass of proton is 2000 time to that of electron

Now due to coulombic force of attraction electronis accelerated towards the proton and the coulombic force is varying continuously due to change in distance between electron and proton

coulombic force is given by

$F = k_C \frac{q_1 q_2}{r^2}$

where

$k_C = \frac{1}{4 \pi \epsilon_0} \approx$

= 9×10⁹ N m² C^-2 (also m F^-1) is the electrostatic constant

so work done by this force when electron attains velocity twice to that of its initial velocity then

work done by force F is given by

(1/2)mv²₂ - (1/2)mv²₁ =

^r F.dr

where m = mass of electron

here v₂ = 2v₁ and v₁ = 3.24 x 10⁵ m/sec

or (3/2)mv²₁ = ^r F.dr = kq² [-1/r]^r_infinite

or (3/2)mv²₁ = kq²/r -----(1)

here, q = 1.6 x 10^-19 C

Now substitute the values in equation (1) and find the value of r

nikhil_ramaswamy

got it, thanks a lot

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