A small plastic ball of mass 6.10 10^-3 kg and charge +0.150 µC?
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. 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 9.20*10^-22 but not the answer
Well, it seems like you're experiencing some electrical tension with this problem. Let me help you lighten the mood and solve it with a touch of humor!
So, we have a small plastic ball hanging out between the plates of a capacitor. It's like that one friend who always finds themselves in the middle of things. Oh, the joys of equilibrium!
Now, let's talk numbers. Assuming that "hang time" angle of 30.0°, we want to find the magnitude of the charge on each plate. Well, isn't that a shocking question?
First things first, let's take a moment to appreciate the ball's mass of 6.10 * 10^-3 kg. It's a tiny, weighty hero in this electrical theater.
Now, we need to consider the area of each plate, which is 0.0150 m2. Imagine these plates as VIP sections, embracing their ballovein guest – charge parties are quite exclusive!
To find the charge on each plate, we'll need to rely on Coulomb's Law, but unfortunately, I'm all out of Coulombs. Maybe you can give it a shot?
Remember, at equilibrium, the force due to gravity must be balanced out by the electric force. So, incorporating all those factors, don't forget to throw in some trigonometry because angles make life more interesting.
Now, put on your thinking cap, get charged up, and calculate the magnitude of the charge on each plate. You've got this!
To find the magnitude of the charge on each plate, we can use the concept of electrostatic equilibrium, along with the force of gravity acting on the plastic ball.
Given data:
Mass of the plastic ball (m) = 6.10 × 10^-3 kg
Charge on the plastic ball (q) = +0.150 µC = +0.150 × 10^-6 C
Angle between the thread and the vertical (θ) = 30.0°
Area of each plate (A) = 0.0150 m^2
First, we can find the gravitational force acting on the plastic ball. Since the ball is in equilibrium, this force is balanced by the electrostatic force between the plates of the capacitor.
Gravitational force (Fg) = m × g
where g is the acceleration due to gravity (approximately 9.8 m/s^2)
Next, we need to find the electrostatic force (Fe) on the plastic ball. The electrostatic force can be found using Coulomb's law:
Fe = k × |q1 × q2| / r^2
where k is the electrostatic constant (9 × 10^9 N·m^2/C^2),
q1 and q2 are the charges on the ball and on the plate, respectively,
and r is the distance between the ball and the plate.
Since the magnitudes of the charges on the ball and on each plate are equal, we can denote the magnitude of the charge on each plate as q. Therefore, the electrostatic force becomes:
Fe = k × q^2 / r^2
In equilibrium, the electrostatic force must balance the gravitational force, so we have:
Fe = Fg
k × q^2 / r^2 = m × g
Rearranging the equation, we can solve for q:
q^2 = (m × g × r^2) / k
Taking the square root of both sides:
q = √[(m × g × r^2) / k]
Now, let's substitute the given values into the equation:
m = 6.10 × 10^-3 kg
g = 9.8 m/s^2
r^2 = A = 0.0150 m^2 (since each plate has this area)
k = 9 × 10^9 N·m^2/C^2
q = √[(6.10 × 10^-3 kg) × (9.8 m/s^2) × (0.0150 m^2) / (9 × 10^9 N·m^2/C^2)]
Calculating this expression gives the magnitude of the charge on each plate.