How Does a Charged Bead Behave in a Capacitor?

[Peetah]

Homework Statement

A small bead of mass m= 34 g and charge of Q=140 mC has a velocity (all in the horizontal direction) of 7 m/s as it enters the gap of a parallel plate capacitor, initially traveling parallel to the plates. The plates are separated by 4 mm. If the bead enters the capacitor 2 mm above the bottom plate, and leaves the capacitor 2 mm above the bottom plate, calculate the potential difference between the two plates.

m = 0.034kg
Q = 0.140 C
v_x=7m/s
y = 0
d = 0.004m

Homework Equations

ma = Fg + qE
V = Ed

The Attempt at a Solution

I was more or less wondering how to conceptually think about this problem. Since the bead leaves the plates at the exact same height it was initially, does that mean that the force of gravity is equal to the electric force? Therefore I can equate

mg = qE

After, I can just sub in V = Ed into the force equation and solve for potential difference.
But with this solution, does that mean that the initial speed is irrelevant?

Thanks

 

[TSny]

Hi Peetah.

Your analysis looks correct to me. The speed is irrelevant.

 

[BvU]

Something strange here. How can a 4 mm bead weigh 34 gram ? And how can it enter 2 mm above the bottom plate and still exit unscathed ? Isn't the top plate smack in the trajectory ?

 

[Peetah]

Its not a 4mm bead, that's the separation distance between the two plates. Bead is point charge

 

[BvU]

I see. And what is its density, approximately, if it is smaller than 4 mm and still weighs 34 gram ?
A bead of r=1 mm has capacitance 9 pF. 140 mC on that gives a potential of a TeraVolt. Lightning!

But: you are not responsible for the credibility of the exercise. And:

Your analysis looks correct to me.

to me too ! Kudos!