A small plastic ball of mass 6.30 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 0.0200 m2. What is the magnitude of the charge on each plate?
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l+l l \ NEGATIVE WALL
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l+l l30deg.\
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To find the magnitude of the charge on each plate, we need to first find the tension in the thread supporting the ball.
The gravitational force acting on the ball is given by the equation F_gravity = mg, where m is the mass of the ball and g is the acceleration due to gravity. In this case, m = 6.30 × 10^(-3) kg and g ≈ 9.8 m/s^2. Therefore, F_gravity = (6.30 × 10^(-3)) × (9.8) = 0.06174 N.
Since the ball is in equilibrium, the tension in the thread must balance the gravitational force. The angle between the thread and the vertical is given as 30.0°. We can use the following trigonometric relation to find the tension:
Tension = F_gravity / cos(30.0°)
Tension = 0.06174 N / cos(30.0°)
Tension = 0.06174 N / 0.866
Tension ≈ 0.07131 N
Now, we can find the magnitude of the charge on each plate using the equation Q = CV, where Q is the charge, C is the capacitance, and V is the potential difference across the plates.
The capacitance of the capacitor can be found using the formula C = ε₀A/d, where ε₀ is the permittivity of free space (ε₀ ≈ 8.85 × 10^(-12) F/m), A is the area of each plate, and d is the separation between the plates. In this case, A = 0.0200 m² and d = 0. The absence of d implies that the plates are touching, so the potential difference across the plates is zero. Thus, Q = CV = 0.
Therefore, the magnitude of the charge on each plate is zero.