A $0.80-\mu mm$-diameter oil droplet is observed between two parallel electrodes spaced $11mm$apart. The droplet hangs motionless if the upper electrode is$20V$more positive than the lower electrode. The density of the oil is $885kg/m^3$.

a. What is the droplet’s mass?

b. What is the droplet’s charge?

c. Does the droplet have a surplus or a deficit of electrons? How many?

We have given that diameter of oil droplet is $0.80-\mu mm$, the space between parallel electrode is $11mm$,the upper electrode is$20V$more positive than the lower electrode if droplet hangs motionless and density of oil is $885kg/m^3$.

We have to find mass of droplet.

Firstly we will convert $\mu m$to$m$,
$0.80\mu m=0.80\times {10}^{-6}m$

We need to find radius of droplet which will be half of its diameter,

$r=\frac{d}{2}=\frac{0.80\times {10}^{-6}m}{2}=0.40\times {10}^{-6}m$=$\frac{4}{3}\times \mathrm{\pi }\times {(0.4\times {10}^{-6}\mathrm{m})}^3=2.68\times {10}^{-19}{\mathrm{m}}^3$

here density is given so we can use the formula of density to find mass of droplet ,

as $\rho =\frac{m}{v}$$\Rightarrow$$m=\rho v$where $v$is volume and$m$is mass of droplet . let it equation $1$

therefore, to find volume , we can use formula of volume of sphere assuming droplet to be spherical

we get ,$v=\frac{4}{3}\mathrm{\pi }{\mathrm{r}}^3$=$\frac{4}{3}\mathrm{\pi }{(0.4\times {10}^{-6}{\mathrm{m}}^3)}^3=2.68\times {10}^{-19}{\mathrm{m}}^3$

so equation $1$becomes ,$m=\rho v=(885kg/m^3)\left(2.68\times {10}^{-19}{m}^3\right)=2.4\times {10}^{-16}kg$

We have given that diameter of oil droplet is $0.80-\mu mm$, the space between parallel electrode is $11mm$, the upper electrode is$20V$more positive than the lower electrode if droplet hangs motionless and density of oil is $885kg/m^3$.

We have to find charge of the droplet.

We are given in the question that droplet hangs motionless, it means electric force becomes equal to weight of the droplet .

electric force will be$F$= $qE$where $E$ is electric field between the electrodes and $q$ is charge of the droplet ( as $E=\frac{F}{q}$)

so , according to given condition , $qE=mg$where $m$is mass on droplet and $g$ is acceleration due to gravity .

therefore, $q=\frac{mg}{E}$, let this be equation 1

so , to find the value of $q$, we need to know the value of $E$as here value of $m$ is given in question and we know the value of $g$.

As potential difference across electrode is given ,

electric field, $E=\frac{V}{d}$$=\frac{17.8V}{0.011m}=16.18.2V/m$

so equation 1 becomes ,$q=\frac{mg}{E}=\frac{\left(2.4\times {10}^{-16}kg\right)\left(9.81m/s^2\right)}{1618.2V/m}=1.5\times {10}^{-18}C$

We have given that diameter of oil droplet is $0.80-\mu mm$ the space between parallel electrode is$11mm$, the upper electrode is$20V$more positive than the lower electrode if droplet hangs motionless and density of oil is $885kg/m^3$.

We have to find that droplet have how many deficient or surplus electrons .

As given in question , upper electrode is positively charged, since both the charges attract each other , that can only be possible if charges are unlike/ opposite .

therefore, droplet must be negatively charged , so it has surplus of electrons ( negatively charged matter /body have excess of electrons )

charge on droplet is $q=-1.5\times {10}^{-18}C$

we also know that charges come in integer multiples of electronic charges units of $e$

total charge isrole="math" localid="1649750170527" $q=Ne$where N is surplus or excess electrons and $e=-1.6\times {10}^{-19}C$that is charge on each electron

we get , $N=\frac{q}{e}=\frac{-1.5\times {10}^{-18}C}{-1.6\times {10}^{-19}C}=9.4~9$

so it has surplus of $9$electrons