JEE Class main Answered

Figure shows the forces acting on each charged sphere connected to a strings of equal length 100 cm.

Each string makes angle 45^o with vertical .

T is tension force acting along the string , that is resolved as Tcos45 vertically and Tsin45 horizontaly as shown in figure.

At equilibrium, vertical component of tension force balances the weight of sphere. Hence we get

T cos45 = mg ........................... (1)

At equilibrium, horizontal component of tension force balances the electrostatic repulsive force F_e between spheres.

Hence we get,

T sin45 = F_e .................(2)

From eqn.(1) and (2) , we get

F_e = mg tan45 = mg ................(3)

( because tan45 = 1 )

Electrostatic force of repulsion between two spheres is

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

Distance between spheres , d = 100√2 cm = 0.1414 m

From eqn.(3) and eqn.(4) , we get