Respuesta :
Answer:
[tex]9.11\times 10^{-15}.[/tex]
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, [tex]\rm e=-1.6\times 10^{-19}\ C.[/tex]
Let the N number of electrons have charge -40 nC, such that,
[tex]\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}.[/tex]
Now, mass of one electron = [tex]\rm 9.11\times 10^{-31}\ kg.[/tex]
Therefore, mass of N electrons = [tex]\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.[/tex]
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 [tex]25\ \rm mg = 25\times 10^{-6}\ kg.[/tex]
Then,
[tex]\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}.[/tex]
It is the required fraction of mass of the droplet.
