Question

In Millikan oil drop experiment a drop of charge Q and radius r is kept at rest between two plates of potential difference of 800 V. Then charge on other drop of radius 2r which is also at reAst with potential difference of 3200 V is
A. $\dfrac{Q}{2}$
B. 2Q
C. 4Q
D. $\dfrac{Q}{4}$

Answer 1

Hint: Find the expressions for the gravitational force and the electrostatic force to equate them in both the cases. Make use of the formula for the electric field between two plates with some potential difference. Substitute the known values and find the charge on the second drop.

Formula used:
${{F}_{g}}=\rho Vg$
where ${{F}_{g}}$ is the gravitational force on a body of volume V and $\rho $ density V. g is acceleration due to gravity.
 ${{F}_{E}}=qE$
where ${{F}_{E}}$ is the electrostatic force on a charge q when placed in an electric field E.
$E=\dfrac{V}{d}$
where E is the electric field between two plates with potential difference V and separated by distance d.

Complete step by step answer:
It is given that both the oil drops are at rest in the two cases. This means that the electrostatic force on the oil drops will balance the gravitational force on the drop.
Therefore, we can write that ${{F}_{E}}={{F}_{g}}$.
The gravitational force on the oil drop of radius r is equal to ${{F}_{g,1}}=\rho {{V}_{1}}g$ …. (i)
In this case, ${{V}_{1}}=\dfrac{4}{3}\pi {{r}^{3}}$.
Substitute these value in (i).
$\Rightarrow {{F}_{g,1}}=\rho \left( \dfrac{4}{3}\pi {{r}^{3}} \right)g=\dfrac{4}{3}\pi \rho g{{r}^{3}}$ …. (ii)
The electrostatic force on these drop is equal to ${{F}_{E,1}}=Q{{E}_{1}}$.
And ${{E}_{1}}=\dfrac{{{V}_{1}}}{d}$.
$\Rightarrow {{F}_{E,1}}=Q\dfrac{{{V}_{1}}}{d}$ ….. (iii).
Now, equate (ii) and (iii).
$\Rightarrow \dfrac{4}{3}\pi \rho g{{r}^{3}}=Q\dfrac{{{V}_{1}}}{d}$ ….. (iv).
Now, the gravitational force on the oil drop of radius 2r is equal to ${{F}_{g,2}}=\rho {{V}_{2}}g$ …. (v)
And in this case, ${{V}_{2}}=\dfrac{4}{3}\pi {{(2r)}^{3}}=\dfrac{32}{3}\pi {{r}^{3}}$.
Substitute these value in (v).
$\Rightarrow {{F}_{g,2}}=\rho \left( \dfrac{32}{3}\pi {{r}^{3}} \right)g=\dfrac{32}{3}\pi \rho g{{r}^{3}}$ …. (vi)
The electrostatic force on these drop is equal to ${{F}_{E,2}}=q{{E}_{2}}$.
And ${{E}_{2}}=\dfrac{{{V}_{2}}}{d}$.
$\Rightarrow {{F}_{E,2}}=q\dfrac{{{V}_{2}}}{d}$ ….. (vii).
Equate (vi) and (vii).
$\Rightarrow \dfrac{32}{3}\pi \rho g{{r}^{3}}=q\dfrac{{{V}_{2}}}{d}$ ….. (viii).
Now, divide (viii) by (iv).
$\Rightarrow \dfrac{\dfrac{32}{3}\pi \rho g{{r}^{3}}}{\dfrac{4}{3}\pi \rho g{{r}^{3}}}=\dfrac{q\dfrac{{{V}_{2}}}{d}}{Q\dfrac{{{V}_{1}}}{d}}$
$\Rightarrow 8=\dfrac{q}{Q}.\dfrac{{{V}_{2}}}{{{V}_{1}}}$
$\Rightarrow q=\dfrac{8Q{{V}_{1}}}{{{V}_{2}}}$
It is given that ${{V}_{1}}=800V,{{V}_{2}}=3200V$.
$\therefore q=\dfrac{8Q(800)}{(3200)}=2Q$.

Hence, the correct option is B.

Note:While performing the Millikan oil drop experiment, we must arrange the set up (the parallel plates) in such a way the electrostatic force on the charged oil drop is in the upward direction as the gravitational force is always in the downward direction.