Calculating Charge Distribution in a Parallel Plate Capacitor

AI Thread Summary
A plastic ball with a mass of 6.95×10^-3 kg and a charge of +0.11 mC is suspended between the plates of a capacitor, creating an equilibrium state at a 30° angle to the vertical. The area of each capacitor plate is 0.0152 m². The electric field (E) is calculated using the formula E = q/(Aε₀), where ε₀ is the permittivity of free space. The mechanical equilibrium of the ball is expressed as the sum of tension, gravitational force, and electric force equating to zero. The discussion highlights the importance of correctly setting up vector components to solve for the charge on each plate effectively.
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A small plastic ball of mass 6.95×10-3 kg and charge +0.11 mC is suspended from an insulating thread and hangs between the plates of a capacitor. The ball is in equilibrium, with the thread making an angle of 30.0° with respect to the vertical. The area of each plate is 0.0152 m2. What is the magnitude of the charge on each plate? (the picture shows the positive plate on the left, the ball leaning 30 deg toward the negative plate, and the negative plate on the right.

With this problem, I solved for E using E=(q)/(AEo). I don't know where to go from here though. I'm confused.
 
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Write the mechanical equilibrium of the ball
\vec{T}+\vec{G}+\vec{F_{el}}=0
onto the horizontal and vertical direction. You now that
G=mg
F_{el}=qE

and E=\frac{Q}{\epsilon_0 A}

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oooo, i set up my vectors wrong in the beginning!
how silly of me, thank you very much for the help.
 
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