Question

three identical positive charge, each of charge 4 micro C are placed at the vertices of an equilateral triangle of side length 10 cm. Calculate the net force acting in any charge at any corner

Answer 1

Let an equilateral triangle is ABC of side length 10cm. each of charge 4 micro C are placed at the vertices of ABC.

let us find net force acting at point A.

force exerted by C on A = $F_{CA}$ = k4$\mu$C × 4$\mu$C/(10cm)²

force exerted by B on A = $F_{BA}$ = k4$\mu$C × 4$\mu$C/(10cm)²

we know, angle between two sides of an equilateral triangle is 60°.

so, net force experienced by A = $\sqrt{F_{CA}^2+F_{BA}^2+2F_{CA}F_{BA}cos60^{\circ}}$

Let $F_{CA}=F_{BA}=F$

so, net force experienced by A = √(F² + F² + 2F² × cos60°)

= √3F

= √3 × k4$\mu$C × 4$\mu$C/(10cm)²

= √3 × 9 × 10^9 × 16 × 10^-12/(0.1)²

= 144√3 × 10^-1

=24.9 N