JEE Class main Answered

Figure shows the position of charges +q on the circular groove of radius r .

These charges are separated so that difference of their angular position is 90^o .

F₁ is the repulsive electrostatic force between charges +q  .

If it is required to put a -Q charge on the groove to keep these +q charges stationary ,

then this -Q charge is placed symmetrically between +q charges as shown in figure.

F₂ is the attarctive force between +q  and -Q charge .

Resolved component ( F₂ cosθ ) balances F₁ and makes the +q charges stationary.

( It is assumed the groove width just equal to the dimension of charges , so that other resolved component

( F₂ sinθ ) is resisted by the groove ) .

Hence , to keep the +q charges stationary , we get

where K = 1/(4πε_o ) = 9 × 10⁹ N m² C^-2 is Coulomb's constant and d is distance between +q and -Q charge.

Distance d is calculated from cosine formula as

hence we get

Hence we get from above expression

Angle θ is calculated as 22.5^{o _.}

Hence we get  ,  Q = 0.317 q