Question

between
1 An oil drop of
mass
3.25 x 10^15kg falls vertically
with uniform velocity through
vertical plates which are 2 cm apart when a pd of
Iooo v is applied to the plates, the drop
moves towards the (-vely) charged plate being
being inclined at 45° to the vertical. Calculte
the charge on the drop and no of electrons
attached to it ?​

Answer 1

From FBD of oil drop, we get,

$\sf{\longrightarrow F\cos45^o=mg\quad\quad\dots(1)}$

$\sf{\longrightarrow F\sin45^o=qE\quad\quad\dots(2)}$

Dividing (2) by (1) we get,

$\sf{\longrightarrow \tan45^o=\dfrac{qE}{mg}}$

$\sf{\longrightarrow qE=mg}$

Since $\sf{E=\dfrac{V}{d},}$

$\sf{\longrightarrow \dfrac{qV}{d}=mg}$

$\sf{\longrightarrow q=\dfrac{mgd}{V}}$

In the question,

• $\sf{m=3.2\times10^{-15}\ kg}$

• $\sf{d=2\times10^{-2}\ m}$

• $\sf{V=1000\ V}$

• $\sf{g=10\ m\,s^{-2}}$

Then, charge on the drop

$\sf{\longrightarrow q=\dfrac{mgd}{V}}$

$\sf{\longrightarrow q=\dfrac{3.2\times10^{-15}\times10\times2\times10^{-2}}{1000}}$

$\sf{\longrightarrow\underline{\underline{q=6.4\times10^{-19}\ C}}}$

and no. of electrons attached to it,

$\sf{\longrightarrow n=\dfrac{q}{e}}$

$\sf{\longrightarrow n=\dfrac{6.4\times10^{-19}}{1.6\times10^{-19}}}$

$\sf{\longrightarrow\underline{\underline{n=4}}}$