Motion in Fields (two parallel plates when voltage is applied)

[jasper10]

1. The problem statement, all variables and given/known data

Two parallel metallic plates are placed horizontally with a separation of 2.2mm. When the voltage is reduced to zero (so that the plates are neutral), the particle (or sphere) is observed to fall with constant velocity. It is observed to travel 0.85 mm in 8.9s.
A new p.d. V volts is now applied to the plates, which causes the particle to move upwards with a constant velocity of 4 x 10^-5 m/s. You may assume for small velocities like this, the frictional force is proportional to the speed.
Calculate the value of V.



Mass of Particle: 3.5 x 10^-15 kg
Charge of Particle: 6.4 x 10^-19 C
g : 9.81ms^-2



2. Relevant equations



3. The attempt at a solution

KE1= Vq = .5mv^2 0= .5 x 3.5 x 10^-15 x (.00085 / 8.9)^2 = 1.596 x 10^-23J
KE2= Vq = .5mv^2 0= .5 x 3.5 x 10^-15 x (4 x 10^-5)^2 = 2.8 x 10^-24

KEtotal= KE1 + KE2 = 1.876 x 10^-23J = Vq
V = 2.9x10^-5 volts

Isn't this value faaar too small?

[jasper10]

Any answers or suggestions are highly appreciated!

[ideasrule]

You forgot to consider air resistance, which robs the particle of energy.

Try calculating this using forces instead of energies. You know that in both cases, the particle is moving at constant velocity, meaning net force = 0.

This problem is extremely easy. You'll see this when you get the answer.

[Redbelly98]

KE1= Vq = .5mv^2 0= .5 x 3.5 x 10^-15 x (.00085 / 8.9)^2 = 1.596 x 10^-23J
KE2= Vq = .5mv^2 0= .5 x 3.5 x 10^-15 x (4 x 10^-5)^2 = 2.8 x 10^-24

These equations don't apply here. This would be true IF the particle started from rest at one plate and had a constant acceleration until it reached the other plate.

Instead, use the fact that
...the frictional force is proportional to the speed
i.e.,
F = -bv
where b is a constant.

[jasper10]

well for the first one (no voltage), v = 9.55 x 10^-5

F = -bv so 9.81 x 3.5 x 10^-15 = -b x 9.55 x 10^-5

is it correct to use 9.81 x 3.5 x 10^-15 as the force? if so then you could find
-b = 9-81x3.5 x 10^-15 /v

then u find F(2) = -b x (4 x 10^-5)

and since F = Vq / D, V can be found. Is this the correct method? But have I taken into account that it is travelling at constant velocity (net force= 0)?

Thanks !!!

[jasper10]

but then i get V to be 49.14 voltswhich is the wrong answer, as it says that when V = 118V, the particle is stationary (not moving up or downwards!)...what have i done wrong?
i.e. what do i use as the force?

[Redbelly98]

You're on the right track. Don't forget that gravity always acts on the particle, whether it is moving upward or downward.

[jasper10]

oh so the total force would be

F = (-b x v) + (mg)= (3.59 x 10^-10 x 4 x 10^-5 ) + (9.81 x 3.5 x 10^-15)

= 4.87 x 10^-14

then, by using F = Vq / D, I get V to be 167V.

This must be the correct answer (at least I hope it is!)

thanks so much!

[Redbelly98]

... I get V to be 167V.

This must be the correct answer (at least I hope it is!)
Looks good (at least that's what I got too) :smile: