we encountered a problem where a charged droplet remains stationary above the earth, the electric field of the earth is given and its asks how many excess electrons are in the droplet.
so i assumed that since there is no movement there has to be no net electric field. so i assumed the electric field of the droplet is also 145 n/c just negative. i used the equation kQ/r^2 = electric field. so i multiplied 145 by the square of the radius and i divided it by the constant k to get the total charge of the droplet. then i divided that by 1.602 E-19 to get the number of electrons present, but i still could not get the right answer, what was i doing wrong?
You are right that there has to be zero net something, however, it isn't the Electric field that cancels to suspend the drop, it is the net Force on the droplet that has to be zero to establish static equilibrium. So, let's think about what forces are exerted on the droplet here. Firstly, there is a force of gravity exerted down on the drop since it is suspended above the Earth. Secondly, because the droplet is charged and is in an Electric field, it will feel a force due to this Electric field. We know that this force must point "up" (ie. oppose gravity), so we can determine from this the sign of the charge. Since the Electric field of the Earth points "down" (radially inward, gravity points radially inward), and the direction of the Electric field is defined as the direction of the force on a positive point charge, the charge of the particle will have to be negative. We can now draw a diagram to make this more clear:
The electric force is Fel=qE. We know E (it is given in the problem), and we are looking for q. For the gravitational force we need to do a bit of work if the mass isn't given directly. Since we know the size (radius) of the drop, we can calculate it's mass from the density, ρ, and volume, V=(4/3)πr3: m=ρV. Then we have all the pieces we need.
It should be noted that the equation for the electric field due to a point charge cannot be used in the way you tried to use it. E=kQ/r2 is the Electric field due to a point charge with charge, Q, at a distance, r, away.
Hope this is helpful to you, if you have follow-up questions don't hesitate to comment on my answer here. Thank-you for being the first to ask a question!
2 comments:
Hi
I have similar question to the question posted above about a water droplet. I solved for a charge by using the above method. I then divided by 1 electron charge (1.6E-19) and got the number of electron charges in the water droplet. My confusion lies in the wording of the specific question. What does "excess" mean? I'm supposed to find how many "excess" electron charges the water droplet has... I know how many electron charges the water droplet has. I also know that in a water molecule there are 2 electrons from H2 and 8 electrons from ). This means that in a mole of a water molecule, there are 6.02E24 electrons. However If i subtract this number from the number of electrons in the charged water droplet... the answer doesn't work. Can you please help explain what I'm doing wrong?
The number of electrons you calculated are the excess charges. All that is meant by this wording is that you do not need to include the number of electrons contained in the water molecules to make a neutral water drop. The negative charge of electrons contained in the water molecules is balanced by the positive charges of the nucleii of those molecules so they do not contribute to a net charge.
Hope that eliminates the confusion.
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