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Questions and Answers
A material has a relative permittivity ($\epsilon_r$) of 5. How does the electric force between two charges placed in this material compare to the force if the charges were in a vacuum, assuming the distance between the charges remains constant?
A material has a relative permittivity ($\epsilon_r$) of 5. How does the electric force between two charges placed in this material compare to the force if the charges were in a vacuum, assuming the distance between the charges remains constant?
The electric force is reduced by a factor of 5.
If the permittivity of a medium is given by $\epsilon = 4.427 \times 10^{-11} CN^{-1}m^{-2}$, what is the relative permittivity ($\epsilon_r$) of the medium? Use $\epsilon_0 = 8.854 \times 10^{-12} CN^{-1} m^{-2}$.
If the permittivity of a medium is given by $\epsilon = 4.427 \times 10^{-11} CN^{-1}m^{-2}$, what is the relative permittivity ($\epsilon_r$) of the medium? Use $\epsilon_0 = 8.854 \times 10^{-12} CN^{-1} m^{-2}$.
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Explain the physical significance of the constant $K = \frac{1}{4 \pi \epsilon _0}$ in the context of electrostatics.
Explain the physical significance of the constant $K = \frac{1}{4 \pi \epsilon _0}$ in the context of electrostatics.
The constant $K$ is a proportionality constant in Coulomb's law that relates the electric force between two point charges to the magnitude of the charges and the distance between them. It reflects the strength of the electrostatic force in a vacuum.
How does increasing the dielectric constant of a medium affect the capacitance of a capacitor, assuming all other factors remain constant?
How does increasing the dielectric constant of a medium affect the capacitance of a capacitor, assuming all other factors remain constant?
A student measures the force between two charges in a medium and finds it to be half the force when the same charges are in a vacuum at the same distance. What is the dielectric constant (K) of the medium?
A student measures the force between two charges in a medium and finds it to be half the force when the same charges are in a vacuum at the same distance. What is the dielectric constant (K) of the medium?
Flashcards
What is K in electrostatics?
What is K in electrostatics?
Electrostatic constant used in Coulomb's Law. Equals approximately 9 x 10^9 Nm^2 C^{-2}.
What is ε₀?
What is ε₀?
Permittivity of free space, representing the ability of a vacuum to permit electric fields. Equals approximately 8.854 x 10^{-12} CN^{-1} m^{-2}.
Force in Air/Vacuum
Force in Air/Vacuum
The force between charges in air or vacuum, using ε₀ in Coulomb's Law.
What is εᵣ (relative permittivity)?
What is εᵣ (relative permittivity)?
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What is the relationship between ε, ε₀, and εᵣ?
What is the relationship between ε, ε₀, and εᵣ?
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Study Notes
- K is equivalent to 1 / (4πε₀) = 9 x 10^9 Nm^2 C^-2
Permittivity of Vacuum
- ε₀ denotes the permittivity of vacuum
- ε₀ = 8.854 x 10^-11 C^2 N^-1 m^-2
- ε₀ is also equivalent to 1 / (36π x 10^9) C^2 N^-1 m^-2
- For air/vacuum, F₀ => q₁q₂ / r^2
Another Medium
- Κ = 1 / (4π ε) = 1 / (4π ε₀ εr)
- ε = ε₀ εr
- εr represents the relative permittivity of a medium
- Dielectric constant is another term for relative permittivity and is denoted by K
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Description
Understand the concepts of permittivity and dielectric constant in electromagnetism. Learn about the permittivity of vacuum (ε₀) and its relation to Coulomb's constant. Explore how these parameters change in different mediums using relative permittivity (εr).
