question_answer

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

Correct Answer: C

Solution :

                In Millikans oil drop experiment, in order that the drop stays stationary force due to electric field should equal to gravitational force. \[\therefore \]  \[F=qE=mg\]                                   ...(i) Also,     \[mass=volume\times density\]                 \[m=\frac{4}{3}\pi {{r}^{3}}\rho \] and         \[E=\frac{V}{d}\] Hence, Eq. (i), becomes                 \[\frac{4}{3}\pi {{r}^{3}}\rho g=q.\frac{V}{d}\] \[\therefore \]  \[\frac{{{V}_{1}}}{{{V}_{2}}}={{\left( \frac{{{r}_{1}}}{{{r}_{2}}} \right)}^{3}}\left( \frac{{{q}_{2}}}{{{q}_{1}}} \right)\] Given, \[{{r}_{2}}=2r,\,{{r}_{1}}=r,\,{{q}_{2}}=2q,\,{{q}_{1}}=q\] \[\therefore \]  \[\frac{{{V}_{1}}}{{{V}_{2}}}={{\left( \frac{r}{2r} \right)}^{3}}\left( \frac{{{q}_{2}}}{{{q}_{1}}} \right)\] \[\therefore \]  \[{{V}_{2}}=4{{V}_{1}}=4V\] (\[\therefore \]\[{{V}_{1}}=V\](given))