A 0.13g plastic bead is charged by the addition of 1.0×1010 excess electrons. What electric field E⃗ (strength) will cause the bead to hang suspended in the air?

Question

A 0.13g plastic bead is charged by the addition of 1.0×1010 excess electrons. What electric field E⃗  (strength) will cause the bead to hang suspended in the air?

theeric · Answer

Since you don't know anything extra about the bead, like its size, you aren't concerned with anything about the bead other than it has a point of charge and a mass.

And what do they mean by suspended? Well, it's floating. Gravity still exerts a force, like normal. But there is an electric field, which tells you the force per charge at its location. This electric field must provide an upwards force to compete  with gravity.

So you have a force from gravity, and a force that you can find with the electric field.

$$F_{gravity}=m\ g$$Where $m$ is the mass of the bead, and $g$ is the acceleration due to gravity.
$$F_{field}=E\ q$$Where the $E$ is the electric field, and $q$ is the charge of the bead.

So, you know $F_{gravity}$, and that $F_{field}$ is opposite to the $F_{gravity}$.

In math terms, $F_{gravity}=-F_{field}$.

More specifically, in math terms (by substitution),$$m\ g=-E\ q\\qquad\Downarrow\-\frac{m\ g}{q}=E$$

Which brings about a question. What is the charge? We know something, the number of excess electrons, which tells us the excess of negative charge.

isaiah.feynman · Answer

theEric the charge can be calculated by using the equation q=ne. Where n is the number of electrons and e is the electric charge (1.602*10^-19).

isaiah.feynman · Answer

E is the elementary charge.

theeric · Answer

Well, one electron is  $-1.602\times 10^{-19}$ of a "coulomb" of charge. That is,$$1[electrons]=-1.602\times 10^{-19} [electrons/coulomb]\ 1[coulomb]\\qquad\qquad\qquad\qquad\qquad\Downarrow\\frac{[electrons]}{-1.602\times 10^{-19}[electrons/coulomb]}=[coulombs]$$
So, you have $1.0\times 10^{10} [electrons]$, so how many coulombs of charge do you have? When I say charge, it's implied to be positive charge. Your answer will be negative because electrons have negative charge.

theeric · Answer

Thanks, @Isaiah.Feynman !

theeric · Answer

So, @11tank , you can find $q$. And then you can calculate $$-\frac{m\ g}{q}=E$$

Please keep in mind that I've tried to keep signs correct. Therefore, you should make sure that the value of $q$ is still negative and that the value of $g$ is negative. You'll probably use $g=-9.8[m/s^2]$, right?

I made the units do that so you don't have to worry about direction otherwise.

Good luck calculating! Reply with any questions you have!

anonymous · Answer

Thank you, @theEric ! This helped me out a lot!

theeric · Answer

I'm glad! Take care!