Analyzing the Forces on the Copper Ball

When the copper ball is immersed in oil and remains suspended in an upward electric field, it means the net force acting on the ball is zero. There are three main forces acting on the ball:

• Weight ($\text{W}$): This is the force due to gravity acting downwards. It is calculated as the mass of the ball multiplied by the acceleration due to gravity ($\text{g}$). Mass is the product of the ball's volume and density.
• Buoyant Force ($\text{F_B}$): This is the upward force exerted by the oil on the ball, according to Archimedes' principle. It is equal to the weight of the volume of oil displaced by the ball. This is calculated as the volume of the ball multiplied by the density of the oil and $\text{g}$.
• Electric Force ($\text{F_E}$): This is the force exerted by the electric field on the charged ball. Since the electric field is acting upwards and the ball is suspended (meaning this force must counteract the net downward forces), the charge on the ball must be positive, causing an upward electric force. The magnitude of the electric force is the product of the charge ($\text{q}$) and the electric field intensity ($\text{E}$).

Equilibrium Condition

For the ball to remain just suspended, the upward forces must balance the downward forces.

[ \text{Upward Forces} = \text{Downward Forces} ]

[ \text{F_B} + \text{F_E} = \text{W} ]

Calculating the Physical Quantities

Let's calculate the volume, weight, and buoyant force. We are given:

• Density of copper ball ($\rho_{ball}$) = 8.0 g/cc = $8.0 \times 1000$ kg/m³ = 8000 kg/m³
• Density of oil ($\rho_{oil}$) = 0.8 g/cc = $0.8 \times 1000$ kg/m³ = 800 kg/m³
• Diameter of the ball (d) = 1 cm = 0.01 m
• Radius of the ball (r) = d/2 = 0.01 m / 2 = 0.005 m
• Electric field intensity (E) = $600\pi$ V/m

We will use the acceleration due to gravity $\text{g} \approx 10$ m/s² for simplicity in calculations, as it leads to one of the provided options.

1. Volume of the copper ball (V):

The ball is a sphere, so its volume is:

[ V = \frac{4}{3}\pi r^3 ]

[ V = \frac{4}{3}\pi (0.005 \text{ m})^3 = \frac{4}{3}\pi (125 \times 10^{-9} \text{ m}^3) ]

2. Weight of the copper ball (W):

[ W = V \times \rho_{ball} \times g ]

[ W = \left(\frac{4}{3}\pi (125 \times 10^{-9} \text{ m}^3)\right) \times (8000 \text{ kg/m}^3) \times (10 \text{ m/s}^2) ]

[ W = \frac{4}{3}\pi (125 \times 10^{-9}) \times 80000 \text{ N} ]

3. Buoyant force (F_B):

[ F_B = V \times \rho_{oil} \times g ]

[ F_B = \left(\frac{4}{3}\pi (125 \times 10^{-9} \text{ m}^3)\right) \times (800 \text{ kg/m}^3) \times (10 \text{ m/s}^2) ]

[ F_B = \frac{4}{3}\pi (125 \times 10^{-9}) \times 8000 \text{ N} ]

4. Electric Force (F_E):

[ F_E = qE ]

[ F_E = q \times (600\pi \text{ V/m}) ]

Solving for the Charge (q)

Using the equilibrium equation $\text{F_B} + \text{F_E} = \text{W}$:

[ \frac{4}{3}\pi (125 \times 10^{-9}) \times 8000 + q (600\pi) = \frac{4}{3}\pi (125 \times 10^{-9}) \times 80000 ]

Rearrange the equation to solve for $\text{q}$:

[ q (600\pi) = \frac{4}{3}\pi (125 \times 10^{-9}) \times 80000 - \frac{4}{3}\pi (125 \times 10^{-9}) \times 8000 ]

[ q (600\pi) = \frac{4}{3}\pi (125 \times 10^{-9}) (80000 - 8000) ]

[ q (600\pi) = \frac{4}{3}\pi (125 \times 10^{-9}) (72000) ]

Divide both sides by $600\pi$:

[ q = \frac{\frac{4}{3}\pi (125 \times 10^{-9}) (72000)}{600\pi} ]

The $\pi$ terms cancel out:

[ q = \frac{\frac{4}{3} (125 \times 10^{-9}) (72000)}{600} ]

[ q = \frac{4 \times 125 \times 10^{-9} \times 72000}{3 \times 600} ]

[ q = \frac{500 \times 10^{-9} \times 72000}{1800} ]

[ q = \frac{500 \times 10^{-9} \times 720}{18} ]

[ q = 500 \times 10^{-9} \times 40 ]

[ q = 20000 \times 10^{-9} ]

[ q = 20 \times 10^{-6} \text{ C} ]

[ q = 2 \times 10^{-5} \text{ C} ]

The calculated charge on the ball is $2 \times 10^{-5}$ C.

Revision Table: Key Parameters and Values

Additional Information: Forces in Fluids and Electric Fields

This problem combines concepts from fluid mechanics (buoyancy) and electrostatics (electric force). Understanding how these forces interact is key to solving such problems.

• Archimedes' Principle: States that the buoyant force on an object submerged in a fluid is equal to the weight of the fluid displaced by the object. This upward force arises because the pressure exerted by the fluid increases with depth.
• Electric Force in a Uniform Field: The force on a charge $\text{q}$ in a uniform electric field $\text{E}$ is given by $\vec{F_E} = q\vec{E}$. If the charge is positive, the force is in the direction of the field; if negative, it's opposite to the field direction. In this problem, the upward electric field exerts an upward force on the positive charge, helping to suspend the ball.
• Equilibrium: When an object is suspended or floats, the net force on it is zero. This is a specific case of Newton's first law of motion. In fluid problems, equilibrium often involves balancing weight, buoyant force, and potentially other forces like tension or, as in this case, electric force.

This problem demonstrates how to apply these fundamental principles to solve a composite physics problem involving multiple forces acting on an object.