How Is Electric Force Balancing a Suspended Charged Ball in a Capacitor?

[stuplato]

A small plastic ball of mass 6.10 10-3 kg and charge +0.150 µC 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° with respect to the vertical. The area of each plate is 0.0150 m2. What is the magnitude of the charge on each plate?
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I tried several times and the most reasonable answer I get is .9715 m but wrong... Thanks in advance

 

[Astronuc]

Please show your work.

This is a statics problem. The electric force = F_e = qE must equal the restoring force for a net force of zero on the ball.

Start by determing the restoring force on the ball, i.e. the gravitational force. Write out the force balance equation.

 

[quicknote]

Based on the given information, we can use the formula for electric force (F) to calculate the magnitude of the charge on each plate:

F = (k * q1 * q2)/d^2

Where k is the Coulomb's constant (9x10^9 N*m^2/C^2), q1 and q2 are the charges on the plates, and d is the distance between the plates.

Since the ball is in equilibrium, we can assume that the electric force on the ball is equal to the weight of the ball (mg), where g is the acceleration due to gravity (9.8 m/s^2).

Therefore, we can set up the following equation:

mg = (k * q1 * q2)/d^2

Substituting the given values, we get:

(6.10x10^-3 kg * 9.8 m/s^2) = (9x10^9 N*m^2/C^2 * q1 * 0.150x10^-6 C)/ (0.0150 m)^2

Solving for q1, we get:

q1 = (6.10x10^-3 kg * 9.8 m/s^2 * 0.0150 m^2)/ (9x10^9 N*m^2/C^2 * 0.150x10^-6 C)

q1 = 0.0098 C

Since the charge on each plate is equal and opposite, the magnitude of the charge on each plate is 0.0098 C. This means that the charge on each plate is +0.0098 C and -0.0098 C, respectively.

I hope this helps to answer your question. Please let me know if you have any further questions.