Question

An oil drop carrying charge 'Q' is held in equilibrium a potential difference of 600V between the horizontal plates. In order to hold another drop of twice the radius of equilibrium, a potential drop of 1600V has to be maintained. the charge on the second drop is?​

Answer 1

at equilibrium condition,

Force experienced due to electric field = weight of oil drop

or, QE = mg , where E is electric field intensity.

we know, potential = electric field × seperation

or, E = V/d

or, QV/d = (volume of oil drop × density ) g

or, QV/d = (4/3πr³ρ)g , where ρ is density of oil drop.

or, Q = (4/3 πρgd) × r³/V

here ,(4/3 πρgd) is a constant term.

so, it is clear that Q is directly proportional to r³ and inversely proportional to V.

i.e., $\frac{Q_1}{Q_2}=\frac{r_1^3}{r_2^3}\frac{V_2}{V_1}$

given, $V_1=600V,V_2=1600,r_2=2r_1$ and $Q_1=Q$

then, Q/$Q_2$ = (1/2)³ × 1600/600

or, Q/$Q_2$ = 1/8 × 8/3 = 1/3

or, $Q_2$ = 3Q

hence, charge on the second drop is 3Q