Question

An oil drop having a mass $4.8\times {10}^{-10}gm$ and charge $2.4\times {10}^{-18}$C, stands still between two charged horizontal plates separated by a distance of 1 cm. If now the polarity of the plates is changed the instantaneous acceleration of the drop is:

( Take $g=10m{s}^{-2}$)

• A. 5 $m{s}^{-2}$
• B. 10 $m{s}^{-2}$
• C. 20 $m{s}^{-2}$
• D. 40 $m{s}^{-2}$

Answer 1

Answer: (C)

20 $m{s}^{-2}$

Force in electric field is given by, $F=qE$
and $F=mg$ (By gravitation)
$So,mg=qE$ (for equilibrium)

$\frac{4.8\times {10}^{-10}\times {10}^{-3}\times 10}{2.4\times {10}^{-18}}=E$

$\Rightarrow E=2\times {10}^6\frac{V}{m}$

Now when polarity is changed force from electric field also comes in the direction of mg.

So, $F_{net}=mg+qE=ma$ (By Newton's Law)
$\Rightarrow g+\frac{q}{m}E=a$
$a=10+\frac{2.4\times {10}^{-18}\times 2\times {10}^6}{4.8\times {10}^{-10}\times {10}^{-3}}$
$=10+10$
$=20m/s^2$