Calculating voltage within a capacitor

1. Aug 20, 2011

NewtonianAlch

1. The problem statement, all variables and given/known data
A small object of mass m carries a charge q and is suspended by a thread between the vertical plates of a parallel-plate capacitor. The plate separation is d. If the thread makes an angle θ with the vertical, what is the potential difference between the plates? (Use m for the mass, g for the acceleration due to gravity, q for the charge, d for the plate separation, and theta for θ as necessary.)

2. Relevant equations

E = V/d
F = mg
F = Eq

3. The attempt at a solution

The net force is 0 since there is no acceleration in any one direction, so the forces acting on the sphere would be equal to each other in magnitude and opposite.

Since the sphere of charge creates an angle with the vertical axis, mg would be equal to the tension (T)*cos θ, so mg = Tcosθ

At the same time there would be a force in the horizontal component due to attraction/repulsion from one of the plates, so that would be F = Eq = Tsinθ (because there is an electric field between the plates)

Whereby E = Tsinθ/q

Since E = V/d, V = Ed

Therefore V = Tsinθ/d

From the vertical component it can be seen that T = mg/cosθ

So now, V = mgsinθ/cosθd => V = mgtanθ/d

Is my reasoning and hence, answer, correct?

2. Aug 20, 2011

ehild

Do not use the same notation for both for gravity and electric force.

Better to say that the horizontal force is due to the electric field between the plates (there is attraction by one plate and repulsion by the other one. )

It is correct up to here.

Wrong. q disappeared, V is inversely proportional to d????

ehild

3. Aug 20, 2011

NewtonianAlch

Hmm, that's weird, I don't know why I did that incorrectly here, when I worked it out by hand I didn't have variables disappearing.

It should in fact be: V = Tsinθd/q

Since T = mg/cosθ

Substituting that leaves us with mgsinθd/cosθq => mgdtanθ/q

Thanks for pointing that out.