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Twopoints 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 chargeand string. Find the maximum velocity of the system of charges A and B. |
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Answer» `2 ms^(-1)`  `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` 
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