A small spherical insulator of mass 8.00*10^-2kg and charge +6.00*10^-7C is hung by a thin wire of negligible mass. A charge of -9.00*10^-7C is held 0.450m away from the sphere and directly to the right of it, so the wire makes an angle with the vertical. Find (a) the angle and (b) the tension in the wire.
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It is on this test
To solve this problem, we can use the principles of electrostatics and the concept of equilibrium.
Let's start with finding the angle:
Step 1: Calculate the electric force between the two charges.
The electric force between two charges can be calculated using Coulomb's Law:
F = (k * |q1 * q2|) / r^2
where F is the electric force, k is the electrostatic constant (9.0 x 10^9 Nm^2/C^2), q1 and q2 are the magnitudes of the two charges, and r is the distance between them.
Let's substitute the given values:
F = (9.0 x 10^9 Nm^2/C^2) * (6.00 x 10^-7 C) * (9.00 x 10^-7 C) / (0.450 m)^2
F ≈ 2.22 x 10^-2 N
Step 2: Analyze the forces acting on the charged sphere.
When the system is in equilibrium, the net force acting on the sphere will be zero. The forces acting on the sphere are:
- Weight (mg), directed downwards
- Tension in the wire, directed along the wire
Since the wire is massless, we can assume that the tension in the wire acts only in the vertical direction.
At equilibrium, the magnitude of the vertical component of the tension force must equal the magnitude of the weight force. Therefore, we can write:
T * cos(θ) = mg
Where T is the tension in the wire, θ is the angle with the vertical, m is the mass of the sphere, and g is the acceleration due to gravity (approximately 9.8 m/s^2).
Step 3: Substitute the known values and solve for θ.
T * cos(θ) = mg
T * cos(θ) = (8.00 x 10^-2 kg) * (9.8 m/s^2)
Now, we have to find the value of T:
Step 4: Analyze the forces acting on the charged particle.
When the system is in equilibrium, the net force acting on the particle will be zero.
The forces acting on the particle are:
- Electric force F, directed towards the left
- Horizontal component of the tension force, directed towards the right
The horizontal component of the tension force can be calculated as:
T * sin(θ) = F
Substituting the known values:
T * sin(θ) = (2.22 x 10^-2 N)
Step 5: Divide the equation obtained in Step 4 by the equation obtained in Step 3 to eliminate T and solve for θ.
(T * sin(θ)) / (T * cos(θ)) = F / (mg)
tan(θ) = (2.22 x 10^-2 N) / ((8.00 x 10^-2 kg) * (9.8 m/s^2))
tan(θ) ≈ 2.85
Now, we can find the value of θ by taking the arctangent of both sides:
θ ≈ arctan(2.85)
After calculating, we find that θ ≈ 71.68° (approximately 71.7°).
Therefore, the angle (θ) is approximately 71.7°.
Now let's calculate the tension in the wire:
Step 6: Substitute the known values into the equation T * cos(θ) = mg and solve for T.
T * cos(θ) = (8.00 x 10^-2 kg) * (9.8 m/s^2)
T * cos(71.7°) ≈ 0.784 N
Therefore, the tension in the wire is approximately 0.784 N.
To recap:
(a) The angle (θ) is approximately 71.7°.
(b) The tension in the wire is approximately 0.784 N.