# Parallel Plate Capacitors

by mayo2kett
Tags: capacitors, parallel, plate
 P: 23 A small plastic ball of mass 6.70×10-3 kg and charge +.0150 uC 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 0.0200 m^2. 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. so i used E=q/(Eo)A which gave me 150000 C/(8.89e-12 c^2/Nm^2)(.0200 m^2)=8.44e17 N/C what do i do from here? how do i incorporate the 30° and and the weight of the ball?? -annie
 PF Gold P: 864 Draw a free body diagram: the ball is in equilibrium. The sum of the forces in each direction must be zero. In the y direction you have the weight and one component of the tension in the string pointing in opposite directions, so they must be equal in magnitude. In the x direction you have the electric force and the other component of the tension pointing in opposite directions, so they must be equal in magnitude. The angle is incorporated in the sin and cos terms in the magnitudes of the components of the tension.
 P: 23 so it would be like this? mass + k|q1||q2|/r^2\sin = E + mg/cos ??? which would give me... (6.70e-3)((((8.99e9)(150000)q2)/(r^2))/sin 30) = (8.436)(.0758)??? i would be solving for q2 right? but what about r^2, what would i make that? -annie
PF Gold
P: 864

## Parallel Plate Capacitors

Not exactly. In this case the electric force is not from a point charge, so Coulomb's law doesn't apply. The electric force on a charge is the charge times the electric field at that point. You can calculate the electric field strength inside the capacitor from
$$\vec{E}=\frac {q_{plate}} {A\epsilon_0}$$
where [itex]q_{plate}[/tex] is the charge on the plates. The electric force is then
$$\vec{F}_{elec}=q_{ball}\vec{E}$$
Equillibrium in the x direction means that the sum of the forces to the left is equal to the sum of the forces to the right:
$$\sum\vec{F}_{left}=\sum\vec{F}_{right}$$
$$q_{ball}\vec{E}=\vec{T}cos(30)$$
where[itex]\vec{T}[/tex] is the tension in the string. To find the tension force, you need to set up the equation for equillibrium in the y direction. In the y direction the sum of the up forces must equal the sum of the down forces. These will be the y component of the tension and the wieght, respectivly. From there just solve for [itex]q_{ball}[/tex] and your done.

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