AP Physics 2: Electric Permittivity - Your Ultimate Guide ⚡

Hey there, future physics pro! Let's dive into the world of electric permittivity. This guide is designed to be your go-to resource, especially the night before the exam. We'll break down the concepts, highlight key points, and make sure you're feeling confident and ready to ace this section. Let's get started!

Coulomb's Law Constant (k)

What is Coulomb's Law Constant?

Let's kick things off with Coulomb's Law constant (k). You'll see this little guy everywhere in electrostatics. It's not just a random number; it's a shortcut that makes our lives easier. Think of it as a 'shorthand' constant, making our calculations cleaner and quicker. It's defined as:

$$k = \frac{1}{4\pi\epsilon_0}$$

The Story Behind 'k'

Here's a fun fact: Oliver Heaviside, a brilliant electrical engineer, noticed that the term 1/(4πϵ₀) kept popping up in his work. Instead of writing it out every time, he cleverly decided to use 'k' as a shorthand. 🤯 This is why you'll see 'k' in many equations related to electric fields and forces.

Coulomb's Law

Now, let's see how 'k' fits into Coulomb's Law, which describes the force between two charged particles:

$$F = k \frac{q_1 q_2}{r^2}$$

Where:

• F is the electrical force.
• q1 and q2 are the charges of the particles.
• r is the distance between the charges.

Permittivity vs. Permeability

Two Key Properties of Space

There are two fundamental properties of space and materials that you need to know: permittivity and permeability.

• Permittivity (ε): Measures how much a material obstructs an electric field. Higher permittivity means weaker electric forces and fields.
• Permeability (μ): Measures how easily a material allows a magnetic field to flow. Higher permeability means more magnetic field flow.

Why They're Both Important

Even though we're focusing on permittivity now, remember that both these concepts are connected. They come together when we talk about light (electromagnetic waves). The speed of light, a fundamental constant, is derived from these two properties. 💡

Electric Permittivity (ε)

What is Permittivity?

Permittivity (ε) is all about how much a material allows an electric field to exist within it. We often measure it relative to the permittivity of free space (vacuum), denoted as ε₀.

Relative Permittivity (εr) or Dielectric Constant (κ)

To measure permittivity in other materials, we use relative permittivity (εr), also known as the dielectric constant (κ). It's calculated as:

$$ε_r = \frac{ε}{ε_0}$$

We'll revisit the dielectric constant when we get to capacitors and circuits. For now, just know that it tells us how much better a material is at storing electrical energy compared to a vacuum.

Permittivity Values

Here's a table of permittivity values for common substances:

Final Exam Focus

Alright, let's focus on what's most important for the exam:

• Coulomb's Law: Understand the relationship between force, charge, and distance.
• Permittivity: Know the difference between permittivity and permeability. Focus on permittivity for now.
• Relative Permittivity: Understand how it relates to the permittivity of free space.
• Units: Make sure you're comfortable with the units for k (N⋅m²/C²) and ε (C²/N⋅m²).

Last-Minute Tips

• Time Management: Don't get bogged down on a single question. Move on and come back if you have time.
• Common Pitfalls: Watch out for unit conversions and make sure you're using the correct formulas.
• Challenging Questions: Break down complex problems into smaller, manageable parts.

Practice Question

Practice Questions

Multiple Choice Questions

1. Two point charges, +q and -q, are separated by a distance r. If the distance is doubled, what happens to the magnitude of the electric force between them? (A) It is doubled. (B) It is halved. (C) It is quadrupled. (D) It is reduced to one-fourth.

2. A material has a relative permittivity of 4. What does this mean for the electric field within the material compared to a vacuum? (A) The electric field is four times stronger. (B) The electric field is four times weaker. (C) The electric field is the same. (D) The electric field is two times weaker.

Free Response Question

Question:

A parallel-plate capacitor is filled with a dielectric material that has a relative permittivity of κ. The capacitor has a charge Q and a potential difference V. The capacitance of the capacitor is C. The separation between the plates is d, and the area of each plate is A.

(a) Derive an expression for the capacitance C in terms of the given parameters, including the permittivity of free space ε₀ and the relative permittivity κ.

(b) If the dielectric material is removed, what happens to the capacitance? Explain your reasoning.

(c) If the charge on the capacitor remains constant after the dielectric is removed, what happens to the potential difference across the capacitor? Explain your reasoning.

Solution:

(a) Derivation of Capacitance:

The capacitance of a parallel plate capacitor with a dielectric is given by:

$$C = \kappa \epsilon_0 \frac{A}{d}$$

• 1 point for correctly using the formula for capacitance of a parallel plate capacitor.
• 1 point for including the dielectric constant κ.

(b) Effect of Removing Dielectric on Capacitance:

If the dielectric material is removed, the relative permittivity κ becomes 1 (since it's now air, which is approximately the same as vacuum). Therefore, the capacitance decreases. The new capacitance C' is:

$$C' = \epsilon_0 \frac{A}{d}$$

• 1 point for stating that the capacitance decreases.
• 1 point for explaining that the relative permittivity decreases.

(c) Effect of Removing Dielectric on Potential Difference:

Since the charge Q remains constant and the capacitance C decreases, the potential difference V across the capacitor must increase. This is because Q = CV, so V = Q/C. If C decreases while Q is constant, V must increase.

• 1 point for stating that the potential difference increases.
• 1 point for explaining the relationship using Q = CV.

Answers to Multiple Choice Questions

1. (D) The force is inversely proportional to the square of the distance.
2. (B) The electric field is weaker due to the higher permittivity.