To determine the dielectric constant of ice at -10 °C, we can use the relationship between the forces acting on charged objects in different media. The dielectric constant, often denoted as ε, is a measure of a material's ability to reduce the electric field between charged objects. In this scenario, we have two charged balls in ice and then in water, and we need to find the dielectric constant of ice.
Understanding the Problem
Initially, the two balls are separated by a distance of 25 cm in ice at -10 °C. When the ice melts to form water at 0 °C, the balls are brought closer together to a distance of 5 cm. The dielectric constant of water at 0 °C is known to be 80. We can use the relationship of forces in different media to find the dielectric constant of ice.
Key Concepts
- Electric Force: The force between two charges in a vacuum is given by Coulomb's law:
F = k * (q1 * q2) / r², where k is Coulomb's constant, q1 and q2 are the charges, and r is the distance between them.
- Dielectric Constant: When a dielectric material is present, the force is reduced by the dielectric constant:
F' = F / ε, where ε is the dielectric constant of the material.
Setting Up the Equations
Let's denote the charges of the balls as q and the dielectric constant of ice as ε_ice. The force between the two charges in ice can be expressed as:
F_ice = (k * q^2) / (ε_ice * r_ice²)
Where r_ice is 25 cm (0.25 m). Thus:
F_ice = (k * q^2) / (ε_ice * (0.25)²)
When the ice melts and the balls are brought to a distance of 5 cm (0.05 m), the force in water can be expressed as:
F_water = (k * q^2) / (ε_water * r_water²)
Where ε_water is 80 and r_water is 5 cm (0.05 m). Thus:
F_water = (k * q^2) / (80 * (0.05)²)
Equating the Forces
Since the charges are the same and we can assume that the forces are equal when the balls are at their respective distances, we can set the two equations equal to each other:
(k * q²) / (ε_ice * (0.25)²) = (k * q²) / (80 * (0.05)²)
We can simplify this equation by canceling out k and q² from both sides:
1 / (ε_ice * (0.25)²) = 1 / (80 * (0.05)²)
Solving for ε_ice
Now, we can rearrange the equation to solve for ε_ice:
ε_ice = (80 * (0.05)²) / (0.25)²
Calculating the values:
- (0.05)² = 0.0025
- (0.25)² = 0.0625
Substituting these values back into the equation gives:
ε_ice = (80 * 0.0025) / 0.0625
ε_ice = 200 / 1 = 200
Final Result
The dielectric constant of ice at -10 °C is approximately 200. This indicates that ice has a significantly higher ability to reduce the electric field compared to air, which is consistent with its molecular structure and properties.
