Answer:
[tex]m = 0.041 kg[/tex]
Explanation:
As we know that the small particle is in equilibrium at an angle of 16 degree with the vertical
so here we can use force balance in vertical and horizontal direction
[tex]T cos\theta = mg[/tex]
[tex]T sin\theta = qE[/tex]
now from above equation we have
[tex]T = \sqrt{(mg)^2 + (qE)^2}[/tex]
also by division of above two equations we have
[tex]\frac{qE}{mg} = tan\theta[/tex]
[tex]qE = mg tan\theta[/tex]
now from above equation again
[tex]T = \sqrt{(mg)^2 + (mg)^2 tan^2\theta}[/tex]
[tex]T = mg sec\theta[/tex]
[tex]0.420 = m(9.81) sec 16[/tex]
[tex]m = 0.041 kg[/tex]
