Question

Oil has a density of 860kg/m^3. An oil droplet is suspended between two plates 1cm apart by adjusting the potential difference between them to 1177 V. When the voltageis removed, the droplet falls and quickly reaches constant speed. It is timed with a stopwatch and falls 3mm in 7.33 s. The viscosity of air is 1.83*10^-5 kg/(m*s).What is the droplet's charge ?

Answer 1

In Part A, you found the droplet's charge q=mg/E.

Knowing the equation for density (p=m/V), we can rearrange it to give mass and then substitute it into q=mg/E.

So, we rearrange our density equation to get m=pV and substitute it into our charge equation (q=mg/E) to get q =(pV)(g)/E.

Now, knowing that we're dealing with a spherical object, we can substitute (4/3)(pi)(r^3) in for the Volume (V) in our equation q=(p)(V)(g)/E, giving us q=p(4/3)(pi)(r^3)(g)/E.

From part C, you know that r = Sqr(9nv/2pg), where n is the viscosity of air.

If we substitute that into our charge equation for r, we get

q = p(4/3)(pi)(Sqr(9nv/2pg))^3(g)/E

And since we know that E = Potential Difference (V) / d, we can substitute this last part in giving us.

q = p(4/3)(pi)(Sqr (9nv/2pg))^3(g)

(V/d)

In the case of your problem. if I were to plug it all in

q = (860kg/m^3)(4/3)(pi) * (Sqr(9(1.83e-5 kg/ms)(3e-3 m/7.33s)/(2(860kg/m^3)(9.81m/s^2)))^3 * (9.81m/s)

(1177V/.01m)

[e represents * 10^]

That should give you 2.40 * 10^-18 C.