What is the weight of the drop inside the capacitor

Question

Ahorizontalplane-platecapacitorischargedtoapotentialdifferenceof3,000V.Theseparation
distanceoftheplatesis10cm.Amercurydropcarryingacharge10-6Cmovesupinsidethe
capacitorwithacceleration2.2m/s2.Determinethemassofthedrop.Whatistheweightofthe
dropinsidethecapacitor?

Answer 1

As given data is V = 3000 v   ,  d = 10 cm = 0.1 m    ,q = 10^(-6) c & a = 2.2 m/s^2....

Now, As we know F = q*E        & V = E*d ;

                           m*a = q*E     i.e,    m*2.2 = 10^(-6) *(V/d)

                   So,   m = 0.0136 kg i.e, 13.6 g.

  Therefore, Weight = m*g =  0.136 ...