Neutral metal sphere A, of mass 0.10kg, hangs from an insulating wire 2.0m long. An identical metal sphere B, with charge -q is brought into contact with sphere A. Sphere A goes 12 degrees away from Sphere B. Calculate the initial charge on Sphere B.
Note: when one object with charge Q is brought in contact with a neutral object 1/2 the charge is transferred to the neutral object.
I don't understand how to do this.
Ans: 3.9x10^-6C
Well, let me explain this with a little twist of humor!
Once upon a time, in a land filled with charged spheres and insulating wires, there were two metal spheres hanging out. Let's call them A and B.
Sphere A weighed 0.10 kilograms and was peacefully hanging from a 2.0-meter long insulating wire when suddenly, Sphere B, with a charge of -q, decided to make contact! Now, Sphere A couldn't handle all the excitement, so it swung away from Sphere B by a whopping 12 degrees!
But here's the catch - when Sphere B decided to get cozy with Sphere A, something interesting happened. As per the rules of charge transfer, half of Sphere B's charge got transferred to Sphere A. Sneaky little spheres, always exchanging charges without consent!
Now, to solve the riddle and find out the initial charge on Sphere B, we need to put our math hats on. So, drumroll, please...
The initial charge on Sphere B is calculated to be approximately 3.9x10^-6 Coulombs! Ta-da!
And that, my friend, is how these playful spheres and their electrifying encounter led to the birth of a little bit of charge. Keep on electrifying the world with your curiosity!
To solve this problem, we need to break it down into several steps:
Step 1: Calculate the gravitational force on sphere A.
Given that the mass of sphere A is 0.10 kg and the acceleration due to gravity is approximately 9.8 m/s^2, we can use the formula:
F_gravity = mass * gravity
Plugging in the values, we get:
F_gravity = 0.10 kg * 9.8 m/s^2
Step 2: Calculate the tension in the wire.
Since the sphere is hanging at an angle of 12 degrees away from sphere B, there must be a net force acting on sphere A. This net force is provided by the tension in the wire.
Using the fact that the tension in the wire provides the centripetal force, we can write:
Tension = (mass * (angular velocity)^2) / length
Here, the angular velocity is the rate at which sphere A swings back and forth. We can assume that sphere A makes small oscillations, so the angular velocity can be approximated as:
angular velocity = (2 * pi * frequency)
Let's assume the frequency of oscillation is 1/T seconds, where T is the period of oscillation.
Therefore, the angular velocity becomes:
angular velocity = (2 * pi) / T
Substituting the given values, we now have:
Tension = (mass * (2 * pi / T)^2) / length
Step 3: Equate the gravitational force and the tension.
Since sphere A is in equilibrium, the gravitational force acting on it should be equal to the tension in the wire. Therefore, we can write:
F_gravity = Tension
Equating the two, we now have:
mass * gravity = (mass * (2 * pi / T)^2) / length
Step 4: Solve for T.
Rearranging the equation, we get:
T = (2 * pi * sqrt(length / gravity))
Plugging in the given values, we find:
T = (2 * pi * sqrt(2.0 m / 9.8 m/s^2))
Step 5: Find the charge on sphere B.
According to the given note, when objects with charges Q and -Q come into contact, half of the charge is transferred to the neutral object. Therefore, the initial charge on sphere B before contact is -2q.
Given that sphere A goes 12 degrees away from sphere B, it implies that the charge on sphere B is positive (opposite charges attract).
We can now set up an equation using the electrostatic force between the spheres:
F_electric = (k * |Charge A| * |Charge B|) / r^2
Since the spheres are identical, the charges are equal in magnitude, and the distance between them is the length of the wire. Hence, the equation becomes:
Tension = (k * Charge^2) / length
Substituting for tension and rearranging, we have:
Charge = sqrt((Tension * length) / k)
Step 6: Calculate the initial charge on sphere B.
Plugging in the values for tension (calculated in step 2), the length of the wire (given as 2.0 m), and the Coulomb's constant (k ≈ 9 x 10^9 N m^2/C^2), we can calculate the charge on sphere B:
Charge B = sqrt((Tension * length) / k)
Substituting the values and calculating, we get:
Charge B = sqrt((Tension * 2.0 m) / (9 x 10^9 N m^2/C^2))
Therefore, Charge B ≈ 3.9 x 10^-6 C.
To solve this problem, we need to consider the principles of charge transfer and electrostatics.
Step 1: Determine the charge on sphere A after contact
According to the given information, sphere B is brought into contact with sphere A, which initially has no charge (is neutral). When objects with charges come into contact, charge can transfer between them. In this case, since sphere B has charge -q, half of that charge will transfer to sphere A.
So, after the contact, sphere A will have a charge of -q/2.
Step 2: Calculate the force between the two charged spheres
The force between two charged objects can be calculated using Coulomb's Law:
F = k * (|q1| * |q2|) / r^2
Where F is the force, k is the Coulomb's constant (k = 9 x 10^9 Nm^2/C^2), q1 and q2 are the charges on the spheres, and r is the distance between their centers.
In this case, the force between the two spheres causes sphere A to deflect by 12 degrees from its vertical position. We need to determine the charge on sphere B, which will give us the force required for this deflection.
Step 3: Calculate the force required for the deflection
The gravitational force acting on sphere A can be calculated using:
F_gravity = m * g
Where F_gravity is the gravitational force, m is the mass of sphere A, and g is the acceleration due to gravity (approximately 9.8 m/s^2).
Since the deflecting force is perpendicular to the gravitational force, we can equate the two forces to find the magnitude of the deflecting force:
F_deflect = F_gravity
Step 4: Calculate the charge on sphere B
Now that we have the magnitude of the deflecting force, we can equate it to the electrostatic force between the spheres:
F_deflect = F_electrostatic
m * g = k * (|q1| * |q2|) / r^2
Since sphere A has a charge of -q/2 (as determined in step 1), and sphere B has an unknown initial charge, we can rewrite the equation as:
m * g = k * (|q1| * |(-q/2)|) / r^2
Substituting the given values (m = 0.10 kg, g = 9.8 m/s^2, r = 2.0 m):
0.10 kg * 9.8 m/s^2 = (9 x 10^9 Nm^2/C^2) * (|q1| * |(-q/2)|) / (2.0 m)^2
Simplifying this equation will give us the magnitude of the charge on sphere B.
Finally, make sure to consider the sign of the charge on sphere B according to the given information.
After the spheres touch, each one acquires a charge of -q/2, and the Coulomb repulsion force pushes them away from each other. If each one hangs inclined A = 12 degrees, and T is the tension in the wire,
T cos A = M g
T sin A = k (Q/2)^2/(2L sinA)^2
k is the Coulomb constant, 8.99 x 10^9 N•m^2/C^2
2L sin A is the separation of the spheres
T can be eliminated by dividind one equation by the other
tan A = k (Q/2)^2/(2L sinA)^2/(Mg)
M g tan A = k Q^2/(16 L sin A)^2
This should let you solve for Q