In Millikans oil drop experiment, an oil drop of radius r and charge q is held in equilibrium between plates of a parallel plate capacitor when the potential difference is V. To keep a drop of radius 2 r with charge 2q in equilibrium between the plates, the potential difference required is :

Question

In Millikans oil drop experiment, an oil drop of radius r and charge q is held in equilibrium between plates of a parallel plate capacitor when the potential difference is V. To keep a drop of radius 2 r with charge 2q in equilibrium between the plates, the potential difference required is : (A) V (B) 2 V (C) 4 V (D) 8 V

Tardigrade Admin · Accepted Answer

In Millikans oil drop experiment, in order that the drop stays stationary force due to electric field should equal to gravitational force.

∴ F = qE = m g \dots

(i)
Also, mass

=

volume

\times

density

m = \frac{4}{3} π r^{3} ρ

and

E = \frac{V}{d}

Hence, Eq. (i), becomes

\frac{4}{3} π r^{3} ρ g = q \cdot \frac{V}{d}

∴ \frac{V _{1}}{V _{2}} = (\frac{r _{1}}{r _{2}})^{3} (\frac{q _{2}}{q _{1}})

Given,

r_{2} = 2 r, r_{1} = r, q_{2} = 2 q, q_{1} = q

∴ \frac{V _{1}}{V _{2}} = (\frac{r}{2 r})^{3} (\frac{q _{2}}{q _{1}})

∴ V_{2} = 4 V_{1} = 4 V (∴ V_{1} = V

(given))

Q. In Millikans oil drop experiment, an oil drop of radius r and charge q is held in equilibrium between plates of a parallel plate capacitor when the potential difference is V. To keep a drop of radius 2r with charge 2q in equilibrium between the plates, the potential difference required is :

V

2 V

4 V

8 V