Suppose a speck of dust in an electrostatic precipitator has 1.00 times 10 to the 12th protons in it and has a net charge of negative 5.00 nanocoulombs. How many electrons does the speck of dust contain? The magnitude of the charge of protons and of electrons is 1.60 times 10 to the negative 19th coulombs.
We can call the number of protons, 1.00 times 10 to the 12th, 𝑁 sub 𝑝. And the charge of the electrostatic precipitator, negative 5.00 nanocoulombs, we can call capital 𝑄. We’re told that both protons and electrons, each have a charge magnitude of 1.60 times 10 to the negative 19th coulombs. We’ll call the charge of a proton 𝑄 sub 𝑝, the charge of an electron 𝑄 sub 𝑒. And we know that 𝑄 sub 𝑝 equals the magnitude of 𝑄 sub 𝑒 which equals 1.60 times 10 to the negative 19th coulombs. We want to solve for the number of electrons that the speck of dust contains. We can call that value 𝑁 sub 𝑒.
If we draw a simple sketch of this precipitator, we’re told that it has both protons and electrons on it and has a net charge of 𝑄 which is negative telling us it has more electrons than protons. Now the total charge 𝑄 on the precipitator is equal to the number of protons on the precipitator times the charge of each proton plus the number of electrons times the charge of each electron. If we rearrange this equation to solve for 𝑁 sub 𝑒, we see it’s equal to the total charge 𝑄 minus the number of protons times the charge of a proton all divided by the charge on an electron 𝑄 sub 𝑒.
We’ve been given all these values and can now plug in for them to solve for 𝑁 sub 𝑒. When we do plug in, notice that we use a positive value for the proton’s charge and a negative value for the charge of the electron.
Calculating this number, we find that, to three significant figures, there are 1.03 times 10 to the 12th electrons on this precipitator. So we see that indeed there are an excess of negative charges which accounts for the overall negative charge of the precipitator.