Question

What fraction of all the electrons in a 25 mg water

dropletmust be removed in order to obtain a net charge of 40
nC?

Answer 1

`9.11* 10^(-15).`

Step-by-step explanation:

The water droplet is initially neutral, it will obtain a 40 nC of charge when a charge of -40 nC is removed from the water droplet.

The charge on one electron,
`\rm e=-1.6* 10^(-19)\ C.`

Let the N number of electrons have charge -40 nC, such that,

`\rm Ne=-40\ nC\\\Rightarrow N=(-40\ nC)/(e)=(-40* 10^(-9)\ C)/(-1.6* 10^(-19)\ C)=2.5* 10^(11).`

Now, mass of one electron =
`\rm 9.11* 10^(-31)\ kg.`

Therefore, mass of N electrons =
`\rm N* 9.11* 10^(-31)=2.5* 10^(11)* 9.11* 10^(-31)=2.2775* 10^(-19)\ kg.`

It is the mass of the of the water droplet that must be removed in order to obtain a charge of 40 nC.

Let it is m times the total mass of the droplet which is
`25\ \rm mg = 25* 10^(-6)\ kg.`

Then,

`\rm m* (25* 10^(-3)\ kg) = 2.2775* 10^(-19)\ kg.\\m=(2.2775* 10^(-19)\ kg)/(25* 10^(-3)\ kg)=9.11* 10^(-15).`

It is the required fraction of mass of the droplet.