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.

What is the tension in the string?

Options

A. `11.25 N`
B. `22.5 N`
C. `45 N`
D. `90 N`

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`