Our Discord hit 10K members! 🎉 Meet students and ask top educators your questions.Join Here!

Numerade Educator

# A charged dust particle at rest in a vacuum is held motion less by an upward-directed $475-\mathrm{N} / \mathrm{C}$ electric field. If the dus particle has a mass of $7.50 \times 10^{-10} \mathrm{kg},$ find (a) the charge or the dust particle and (b) the number of electrons that must be added to neutralize it.

## a. $1.55 \times 10^{-11} \mathrm{C}$b. $9.67 \times 10^{7}$

### Discussion

You must be signed in to discuss.
ZU

Zeynep U.

March 25, 2020

A helium nucleus of mass

### Video Transcript

for this problem. On the topic of electric forces and fields, we're told that they charged US particle, which is addressed in a vacuum, is held motionless by electric field with a given magnitude and pointed upward. If we have the mass of the dust particle, we want to find the charge in the dust particle as well as the number of electrons that must be added to neutralize it. So if we draw a force diagram on the dust particle, we know there's an electric field pointed upward, which means there's an electric force pointing upward. We'll call it F E, and this must balanced the weight of the dust particle or F g the force due to gravity since the particle is held addressed. So since these two forces are balanced, we can see F E is equal to F G. The electric force we know is Q times E, where Q is the magnitude of the charge of the dust particle and this must equal the weight, which is M. Times G, where M is the mass of the dust particle so we can rearrange this equation and solve for Q, so Q is the weight of the dust particle mg divided by the electric field strength e and this is the mass of the dust particle 7.5 times 10 to the minus 10 kg times G, which is 9.8 m, the square second, divided by the electric field strength, which is 400 and 75 Newtons per column. Since all our values and sa units we get, he charged Q on the dust particle and S I units to be 1.547 times 10 to the minus 11, which we can write as one 0.55 keeping our significant figures times 10 to the minus 11 columns. So now that we have the charge in the dust particle, we can use this to calculate the number of electrons required to neutralize the charge. So this number of electrons N is the total charge 1.55 times 10 to the minus 11. Hollande's divided by the charge of a single electron, which is 1.6. So we'll use the value 1.5 for seven, divided by 1.6 times 10 to the minus 19 columns, which is the charge on a single electron and we get the value of the number of electrons to be nine 0.67 times 10 to the power seven electrons.

University of Kwazulu-Natal