Question

What fraction of all the electrons in a 25 mg water

Answer 1

Answer:

$9.11\times 10^{-15}.$

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, $\rm e=-1.6\times 10^{-19}\ C.$

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

$\rm Ne=-40\ nC\\\Rightarrow N=\dfrac{-40\ nC}{e}=\dfrac{-40\times 10^{-9}\ C}{-1.6\times 10^{-19}\ C}=2.5\times 10^{11}.$  

Now, mass of one electron = $\rm 9.11\times 10^{-31}\ kg.$

Therefore, mass of N electrons = $\rm N\times 9.11\times 10^{-31}=2.5\times 10^{11}\times 9.11\times 10^{-31}=2.2775\times 10^{-19}\ kg.$

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 $25\ \rm mg = 25\times 10^{-6}\ kg.$

Then,

$\rm m\times (25\times 10^{-3}\ kg) = 2.2775\times 10^{-19}\ kg.\\m=\dfrac{2.2775\times 10^{-19}\ kg}{25\times 10^{-3}\ kg}=9.11\times 10^{-15}.$

It is the required fraction of mass of the droplet.