Two points charges are placed on x-axis as shown, a = 1 cm, `q = 1muC` , mass of each particle `m=6g`. The charges are tied at the end of an inextensible string of length 2a. The whole system is free to move on a horizontal frictionless surface.

Now suppose a third charge of equal value `+q` is fixed at the point `(a//2,0)` and the system of charges A and B is released. There is no friction between third charge and string. Find the maximum velocity of the system of charges A and B.

Options

A. `2 ms^(-1)`
B. `10 ms^(-1)`
C. `10cm s^(-1)`
D. `5 ms^(-1)`

Answer

B

Explanation

`T=F=(kq^(2))/((2a)^(2))=(9xx10^(9)xx10^(-12))/(4xx10^(-4))=22.5`
When velocity is maximum third cahrge will be centre of A and B

`(KE+PE)_(i)=(KE+PE)_(f)`
`0+(kq^(2))/(a//2)+(kq^(2))/(3a//2)=(1)/(2)(2m)v^(2)+2(kq^(2))/(a)`
or `v=sqrt((2)/(3)(kq^(2))/(ma))=sqrt((2)/(3)xx(9xx10^(9)xx10^(-12))/(6xx10^(-3)xx10^(-2)))=10ms^(-1)`
When third charge is at center of A and B, net force due to third charge on A and B is zero. So acceleration of system of A and B is zero. it means individual acceleration of A and B is also zero.
`T=F_(1)+F_(2)=(kq^(2))/(a^(2))+(kq^(2))/((2a)^(2))=112.5N`