Question

Three protons and two positrons afe held by some external means.Protons are present on two opposite corners and one more proton on the centre of square and positrons are present on two remaining corner of the square of side ao.Now external means is removed and the only interaction left is electromagnetic interaction between the particles.Kinetic energy of all three protons after sufficiently long time will be

Answer 1

@ganeshie8

Answer 2

Sounds like physical science and NOT English

Answer 3

Yup ! It is!! Since i have posted questions there so i don't have other choice left so did it here!

Answer 4

ok well your problem is that I am not even seeing a question. What are you asking @samigupta8

Answer 5

Sorry bt i never asked you to help i called ganeshie8 for that purpose

Answer 6

very well, good luck

Answer 7

lol, once again, good luck because clearly your friend is not coming unfortunately @samingupta8

Answer 8

Is the answer 9.8/a nJ ?

Answer 9

Options are
A).3Ke^2/5ao(4+5√2)
B).5Ke^2/√2ao
C).2Ke^2/5ao(4+5√2)
D).9Ke^2/2ao

Answer 10

http://www.wolframalpha.com/input/?i=k*e%5E2*(+(1%2Fa+%2B+1%2Fa)+%2B(1%2Fa+%2B+1%2Fa+%2B+1%2F(a*sqrt(2)))+%2B+(1%2F(a*sqrt(2)%2F2)+%2B+1%2F(a*sqrt(2)%2F2)+%2B+1%2F(a*sqrt(2)%2F2)+%2B+1%2F(a*sqrt(2)%2F2))++)

Answer 11

Exactly!! @ganeshie8 i also got the same ans..
Ke^2/ao[4+4√2+1/√2] =3/2mv^2

Answer 12

Bt one more thing .
We considered only potential energy of the protons with all other similar protons and positrons.
And we didn't calculate the total potential energy of the system as such..
Is it right??

Answer 13

Sorry i didn't mention an important thing in this question
(Given that mass of positron is negligible in comparison to mass of proton)..

Answer 14

I got it!!! Thanks@ganeshie8

Answer 15

Hey @samigupta8

Answer 16

may i see your work ?

Answer 17

Sure!!

Answer 18

Since i mentioned in my post that mass of positron is negligible in comparison to the mass of proton that implies that when external forces are removed then positron will have a very high kinetic energy and which would cause them to just move out of the system like anything!!..
So we will b having just 3 protons in the system under consideration..
So we write the potential energy of this system and that would trn out to be the kinetic energy of the three protons at infinite time

Answer 19

@ganeshie8 did u see it?

Answer 20

do you mean like this ?
http://www.wolframalpha.com/input/?i=k*e^2*%28+%281%2F%28a*sqrt%282%29%29%29+%2B+%281%2F%28a*sqrt%282%29%2F2%29+%2B+1%2F%28a*sqrt%282%29%2F2%29+%29++%29

Answer 21

Yaa!!

Answer 22

Makes sense... thanks :)

Answer 23

I didn't consider that at the first attempt though...
And ended up with an indeed wrong ans ...i came up with option A for the very first attempt(here i considered that all of them have same velocity)..and even same mass..lol ..

Answer 24

yeah this was a bit tricky haha
tell me if i understand the solution correctly :

Answer 25

The mass of positrons is negligible compared to the mass of protons. Therefore, immediately after the external means are removed, these positrons get accelerated far away immediately. So the positrons can no longer affect the mechanical energy of the 3-protons system.

Answer 26

Exactly!!
When they are not included in the system then they don't even worth being a part of the system's total energy...

Answer 27

Awesome! thanks for taking time to explain :)

Answer 28

No problem!!

Answer 29

Shall i ask you a question pls?

Answer 30

Would you help in ydse question??