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Suppose a speck of dust in an electrostatic precipitator has 1×1012 protons in it and has a net charge of −5.2 nC (a very large charge for a small speck). How many electrons does it have?

Respuesta :

Explanation:

The given data is as follows.

         [tex]q_{net}[/tex] = -5.2 nC = [tex]-5.2 \times 10^{-9} C[/tex]  

          charge on an electron (e) = [tex]1.6 \times 10^{-19} C[/tex]

Formula to calculate number of electrons is as follows.

               Q = ne

or,           n = [tex]\frac{Q}{e}[/tex]

Putting the given values into the above formula as follows.

                n = [tex]\frac{Q}{e}[/tex]

                   = [tex]\frac{-5.2 \times 10^{-9} C}{1.6 \times 10^{-19} C}[/tex]

                   = [tex]-3.25 \times 10^{10} C[/tex]

We know that when total charge is zero then number of protons is equal to the number of electrons.

That is,      [tex]n_{p} = n_{e}[/tex]

So, total number of electrons present are as follows.

              [tex]n_{e} = 1 \times 10^{12} + 3.25 \times 10^{10}[/tex]

                          = [tex]103.25 \times 10^{10} C[/tex]

Therefore, we can conclude that given speck of dust has [tex]103.25 \times 10^{10}[/tex] electrons.

So, the required number of electrons is,

[tex]1.03\times 10^{12}[/tex]

A number of electrons:

The number of electrons in a neutral atom is equal to the number of protons. The number of protons in the nucleus of the atom is equal to the atomic number (Z). The number of neutrons is equal to the difference between the mass number of the atom (M) and the atomic number (Z). The mass number of the atom (M) is equal to the sum of the number of protons and neutrons in the nucleus.

Given, the required number of electrons is,

Net charge=-5.2 nC

The total charge of the body,

So, the total charge on a body is equal to the difference in the number of protons and electrons multiplied by the modulus of fundamental charge that a single proton or electron contains. Then from above, we have,

[tex]N_c=N_p-\frac{Q}{q_c}[/tex]

[tex]1\times1012[/tex] protons for [tex]N_p[/tex]  and [tex]-5.2 \times10^{-9}C[/tex] for the total charge Q is,

[tex]1\times10^{12}-\frac{-5.2\times 10^{-9}}{-1.6\times 10^{-19}}=1.03\times 10^{12}[/tex]

Learn More about the topic number of electrons: https://brainly.com/question/933198