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# electrostatics-millikan oil drop watch

1. I am struggling to complete this problem:
A uniform electric field 500 kV m–1 acts vertically downwards. A small oil drop carrying charge –8e falls with a uniform speed of 200 μm s–1. If the charge is increased to –11e, it moves upwards with a speed of 100 μm s–1. Find
(i) the mass of the drop;
(ii) the charge it would carry if it were suspended at equilibrium in the field.

My mass for part i) is half what it should be. I haven't used the velocities given so I know my reasoning isn't quite right, but this is what I have done. Because the speed is uniform, a=0, so my weight force must equal my electric force acting opposite direction. If i choose the -8e charge, my mass is almost half out. Given the correct mass, if I do the same procedure for b) i.e. qe=mg, then q comes out correct...what method should I be using for a)? is there a way of using KE since velocities are given?

thanks,
w
2. eV =1/2mv^2 You don't know voltage V and you don't know mass of oil drop m.
So you can derive two equations using the different values for v and solve for the two unknowns V and m.
3. (Original post by xlaser31)
eV =1/2mv^2 You don't know voltage V and you don't know mass of oil drop m.
So you can derive two equations using the different values for v and solve for the two unknowns V and m.
ty so much
4. On second thoughts this is much more complicated than I first expected.
My previous comment would not work because the m's will cancel and
due to the fact the electrons are glued to the oil drop may invalidate the eV equation.
http://vlab.amrita.edu/?sub=1&brch=195&sim=357&cnt=1
Its a very advanced question (or a very bad question). Well beyond A-level.

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