Question

three charged particles each having a charge q are kept fixed at the vertices of an equilateral triangle of side lenght l . the net electrostatic force on any one of the charged particles is​

Answer 1

three charged particles each having a charge q are kept fixed at the vertices of an equilateral triangle of side lenght l.

Let an equilateral triangle is ABC,

then, force exerted by B on C, F = kq²/l²

force exerted by A on C, F = kq²/l²

see figure, it is clear that both forces are made angle 60° with each other.

so, resultant force experienced by C = √{F² + F² + 2F²cos60°}

= √{2F² + 2F²cos60°}

=√{2F² + F²}

= √3F

putting F = kq²/l²

so, resultant force = √3kq²/l²

direction of resultant force on C :

in vector form, force exerted by B on C = kq²/l² î

in vector form, force exerted by A on C = (kq²/l²)cos60° î - (kq²/l²)sin60° j

so, resultant force, Fnet = (kq²/l²)(1 + cos60°)î - (kq²/l²)sin60° j

= (3kq²/2l²) î - (√3kq²/2l²) j

so, angle made by Fnet with x - axis = tan^{-1}{√3kq²/2l²}/{3kq²/2l²}

= tan^{-1}{1/√3}

= π/6

Answer 2

☆$\red {HELLO\:DEAR}$☆

Opposite charges attract each other and same charges repel each other.

It can be observed that particles 1 and 2 both move towards the positively charged plate and repel away from the negatively charged plate.

Hence, these two particles are negatively charged.

It can also be observed that particle 3 moves towards the negatively charged plate and repels away from the positively charged plate.

Hence, particle 3 is positively charged.

The charge to mass ratio (emf) is directly proportional to the displacement or amount of deflection for a given velocity.

Since the deflection of particle 3 is the maximum, it has the highest charge to mass ratio.

☆$\green {VISHU\:PANDAT}$☆

☆$\blue {FOLLOW\:ME}$☆