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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(213)
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=F22F1cos30 (Horizontal components are cancelled)

Let charges are revolving at the speed of v.
Then
Fcentripetal=mv2R
Fc=mv2R
2kq2R22.kq23R2.32=mv2R
23kq2kq23R2=mv2R
kq2R[213]=mv2
v= kq2mR[213]

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

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