Question Part Points Submissions Used Two hard rubber spheres, each of mass m =

Question

Question Part Points Submissions Used Two hard rubber spheres, each of mass m = 14.4 g, are rubbed with fur on a dry day and are then suspended with two insulating strings of length L = 5.20 cm whose support points are a distance d = 3.30 cm from each other as shown in the figure below. During the rubbing process, one sphere receives exactly twice the charge of the other. They are observed to hang at equilibrium, each at an angle of ? = 11.1
Question Part Points Submissions Used
Question Part Points Submissions Used
Question Part Points Submissions Used
Question Part Points Submissions Used
Question Part Points Submissions Used Two hard rubber spheres, each of mass m = 14.4 g, are rubbed with fur on a dry day and are then suspended with two insulating strings of length L = 5.20 cm whose support points are a distance d = 3.30 cm from each other as shown in the figure below. During the rubbing process, one sphere receives exactly twice the charge of the other. They are observed to hang at equilibrium, each at an angle of ? = 11.1 Two hard rubber spheres, each of mass m = 14.4 g, are rubbed with fur on a dry day and are then suspended with two insulating strings of length L = 5.20 cm whose support points are a distance d = 3.30 cm from each other as shown in the figure below. During the rubbing process, one sphere receives exactly twice the charge of the other. They are observed to hang at equilibrium, each at an angle of ? = 11.1 Question Part Points Submissions Used Two hard rubber spheres, each of mass m = 14.4 g, are rubbed with fur on a dry day and are then suspended with two insulating strings of length L = 5.20 cm whose support points are a distance d = 3.30 cm from each other as shown in the figure below. During the rubbing process, one sphere receives exactly twice the charge of the other. They are observed to hang at equilibrium, each at an angle of ? = 11.1 degree with the vertical. Find the amount of charge on each sphere. (Enter your answers from smallest to largest.)

Explanation / Answer

Since the system is in equilibrium, we know that

The electrical force of repulsion must be balancing the gravitational force of attraction.

So,

let the gravitataional force of attraction be Fg

let the electrical force of repulsion be Fe

then, from Newton's law of universal gravitation we have

Fg = G m1 m2 / r^2

From Coulomb's law we have

Fe = K q1 q2 / r^2

Putting all this together we get

Fe = Fg

K q1 q2 / r^2 = G m1 m2 / r^2

We are told that q1 is twice q2. So, if we let q1 = C coulombs , q2 = 2C coulombs

K * 2C * C / r^2 = G * m1 * m2 / r^2

But m1 = m2 so

K 2C^2 / r^2 = G m^2 / r^2

C^2 = G m^2 r^2 / 2K r^2

C^2 = G m^2 / 2K

C =

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