    Question

# Three equal point charges with charge +q each are moving along a circle of radius R and a fixed point charge −2q is placed at the centre of the circle. If +q charges are revolving at constant speed around −2q as shown, then the speed of charges are (Assume that charges (+q) form vertices of equilateral triangle) A
kq23Rm
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B
 kq2Rm(213)
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C
kq2Rm(2+1)
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D
kq22Rm
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Solution

## The correct option is B  ⎷kq2Rm(2−1√3)We can apply coulomb's law in this case, as relative distance between charges remains same during motion . From the symmetry of the figure, we can see that triangle ABC is an equilateral triangle. Thus, each angle is 60∘. Each side of ΔABC is, (a=AB=BC=CA) ⇒a=BC=Rcos30∘+Rcos30∘ ∴BC=2Rcos30∘=√3R ⇒a=√3R Again from symmetry, the repulsive force experienced by charge at A from charges at corner B and C will be same. The force experienced by charge at A due to charge at B or C is F1=k(q)(q)(√3R)2=kq23R2 The force experienced by charge A due to charge at center is F2=kq(2q)R2=2kq2R2 ⇒ Net force on charge at A is toward centre. Fc=F2−2F1cos30∘ (Horizontal components are cancelled) Let charges are revolving at the speed of v. Then Fcentripetal=mv2R ⇒Fc=mv2R ⇒2kq2R2−2.kq23R2.√32=mv2R ⇒2√3kq2−kq2√3R2=mv2R ⇒kq2R[2−1√3]=mv2 ∴v= ⎷kq2mR[2−1√3] Why this question ?Tip: It challenges you to simplify the problem byusing symmetry of circular path and then apply equationof dynamics towards the centre of circle  Suggest Corrections  4      Similar questions  Explore more