Very Difficult Electrical Field Question!!!??

Question

Question: While you read this question look at this diagram http://i3.tinypic.com/3zho01t.jpg...

A small plastic ball with a mass of 6.50 x 10^-3 kg and with a charge of +0.150 uC is suspended from an insulating thread and hangs between the plates of a capacitor (see the drawing). the ball is in equilibrium, with the thread making an angle of 30.0 degrees with respect to the vertical. the area of each plate is 0.0150 m^2. what is the magnitude of the charge on each plate?

My Approach: I tried to find q using Q = mg/E, but it ended up in a mess since I was looking for the magnitude. The area is throwing me off...I don't know how to utilize it. In general, this is a very tough question! would love all the help i can get. I've tried everything and even a little hint wont help! someone show me the whole solution b/c i give up.

(note: i will award a best answer)

Answer 1

We know q of our plastic ball and we know the mass of the ball and the area of the plate as well as the angle of the string. One thing to note is the electric field in the middle of the plates of a paralell plate capicitor is constant. We want to find this electric field magnitude as it relates to the area of the plates and charge of the plates. The electric field can be written as:

E = sigma/epsilon0 where sigma is the charge density (charge per unit area) of a single plate. (see my first reference)

What is this electric field strength is the next question. You are definitely on the right track with your approach to finding this. Since the ball is in equilibrium, there in no net force acting on it. Therefore the tension in the string must be equal in magnitude and opposite in direction to the vector sum of the electric field force and the gravitational force.

Fg = m g (in - y direction)
Fe = q E (in + x direction)
T is tension in the string

The string is at a 30 degree angle to the verticle which is what allows us to solve the problem.

T cos (30) = Fg
T sin (30) = Fe

Solve both for T to get an expression of Fe and Fg

Fg / cos(30) = T = Fe / sin (30)

Sin (30) = 1/2 and cos (30) = sqrt(2)/2

m g / (sqrt(2)/2) = q E / 1/2

2 m g / sqrt (2) = 2 q E

m g / sqrt (2) = q E

E = m g / (sqrt (2) * q)

So we now know E, awesome!

Remember that E is related to the charge density on the plates.

E = sigma/epsilon0

sigma = Q / A

E = Q / (A epsilon0) = m g / (sqrt (2) q)

So we can solve for charge on the plates as:

Q = A*epsilon0*m*g / (sqrt (2)*q), where eqsilon0 is the permittivity of free space constant

I hope this helps out.

Answer 2

first of all with the given weight you actually can determine the thread's tension.with the tension get the horizontal component of the thread which is same with the electric force
then with the force obtained, you can determine it's separation between the plates and thus you can determine the charge of the plate by using equation
V=Ed
that's ali, i hope you could find the answer with this hints.

Answer 3

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