Electric Field Calculation: Homogeneous Sphere With Charge Density

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Hey guys! Let's dive into a classic physics problem today: calculating the electric field generated by a uniformly charged sphere. This is a super important concept in electromagnetism, and understanding it will help you tackle a wide range of problems. We'll break it down step-by-step, so don't worry if it seems intimidating at first. We've got a sphere with some charge packed inside, and we need to figure out how that charge creates an electric field around it.

Problem Statement

We have a sphere that's made of a homogeneous, non-conducting material. This means the material is the same throughout, and it doesn't let electric charge move freely. The sphere's radius, R, is 14.0 cm, and its volume is approximately 1.1494 x 10^4 cm³. Now, this sphere isn't just empty space; it's filled with electric charge! The charge is spread evenly throughout the sphere, with a constant volume charge density, ρ, of 2.262 x 10⁻³ C/m³. Our mission, should we choose to accept it, is to calculate the electric field (E) created by this charged sphere. We need to find both the magnitude (how strong it is) and the direction (where it's pointing) of this electric field.

Understanding the Key Concepts

Before we jump into the calculations, let's make sure we're all on the same page with the key concepts:

  • Electric Field: Imagine a region around a charged object. The electric field is a way of describing the force that another charge would feel if it were placed in that region. It's a vector quantity, meaning it has both magnitude and direction.
  • Volume Charge Density (ρ): This tells us how much electric charge is packed into a given volume. It's measured in Coulombs per cubic meter (C/m³). In our case, the charge is spread uniformly, so ρ is constant throughout the sphere.
  • Gauss's Law: This is the superhero of electromagnetism! Gauss's Law provides a powerful way to calculate electric fields, especially in situations with symmetry. It relates the electric flux through a closed surface to the enclosed electric charge. In simpler terms, it tells us how the electric field "flows" out of a surface depending on the amount of charge inside.
  • Gaussian Surface: This is an imaginary closed surface that we use in conjunction with Gauss's Law. The clever choice of a Gaussian surface can make our calculations much easier. For spherical symmetry, a spherical Gaussian surface is usually the way to go.

The Strategy: Gauss's Law to the Rescue!

We're going to use Gauss's Law to solve this problem because it's perfectly suited for situations with spherical symmetry, like our charged sphere. Here's the game plan:

  1. Choose a Gaussian surface: We'll pick a spherical Gaussian surface centered on our charged sphere. We'll consider two cases: one where the Gaussian surface is inside the charged sphere (r < R), and one where it's outside the charged sphere (r > R).
  2. Calculate the enclosed charge (Q_encl): This is the amount of charge contained within our Gaussian surface. It'll depend on whether we're inside or outside the sphere.
  3. Apply Gauss's Law: Gauss's Law states that the electric flux through the Gaussian surface is proportional to the enclosed charge. Mathematically, it's written as: ∮ E ⋅ dA = Q_encl / ε₀ Where:
    • ∮ E ⋅ dA is the electric flux (the integral of the electric field over the surface area)
    • Q_encl is the enclosed charge
    • ε₀ is the permittivity of free space (a constant value)
  4. Solve for the electric field (E): We'll use the symmetry of the problem to simplify the integral and solve for the magnitude of the electric field.
  5. Determine the direction: The electric field will point radially outward if the charge is positive (as in our case) and radially inward if the charge is negative.

Calculating the Electric Field Inside the Sphere (r < R)

Alright, let's roll up our sleeves and start crunching some numbers! We'll begin by finding the electric field inside the charged sphere. This is where things get a little interesting because not all the charge is "enclosed" by our Gaussian surface.

1. Choosing the Gaussian Surface

Imagine drawing an imaginary sphere (our Gaussian surface) inside the charged sphere, with a radius r that's smaller than the sphere's radius R. This Gaussian surface is centered on the same point as our charged sphere. Because of the spherical symmetry, this is the perfect choice for applying Gauss's Law.

2. Calculating the Enclosed Charge (Q_encl)

This is the crucial part. We need to figure out how much charge is actually inside our Gaussian sphere. Since the charge is distributed uniformly, we can use the volume charge density to help us.

The volume of our Gaussian sphere is (4/3)πr³. The enclosed charge is simply the volume charge density multiplied by this volume:

Q_encl = ρ * (4/3)πr³

3. Applying Gauss's Law

Now, let's bring in the big guns: Gauss's Law. We know that:

∮ E ⋅ dA = Q_encl / ε₀

Because of the symmetry, the electric field E will point radially outward (since the charge is positive), and its magnitude will be constant over the surface of our Gaussian sphere. This means the angle between E and the area vector dA is 0, and E ⋅ dA simplifies to E dA. The integral then becomes:

E ∮ dA = Q_encl / ε₀

The integral ∮ dA is simply the surface area of our Gaussian sphere, which is 4πr². So we have:

E * 4πr² = Q_encl / ε₀

4. Solving for the Electric Field (E)

We can now substitute our expression for Q_encl and solve for E:

E * 4πr² = [ρ * (4/3)πr³] / ε₀

Notice that 4πr² cancels out on both sides, leaving us with:

E = (ρ * r) / (3ε₀)

This is our electric field inside the sphere! It tells us that the electric field increases linearly with the distance r from the center of the sphere.

5. Determining the Direction

As we mentioned earlier, the electric field points radially outward because the charge is positive. So, inside the sphere, the electric field lines radiate outwards from the center.

Calculating the Electric Field Outside the Sphere (r > R)

Now, let's tackle the electric field outside the charged sphere. This is a slightly different scenario because our Gaussian surface will now enclose the entire charged sphere.

1. Choosing the Gaussian Surface

We'll again choose a spherical Gaussian surface centered on the charged sphere, but this time, the radius r of our Gaussian surface will be larger than the sphere's radius R.

2. Calculating the Enclosed Charge (Q_encl)

Here's the key difference: since our Gaussian surface is outside the sphere, it encloses all of the charge. So, Q_encl is simply the total charge of the sphere, which we can calculate using the volume charge density and the sphere's volume:

Q_encl = ρ * (4/3)πR³

Notice that we're using the sphere's radius R here, not the Gaussian surface's radius r.

3. Applying Gauss's Law

We apply Gauss's Law just like before:

∮ E ⋅ dA = Q_encl / ε₀

Again, due to symmetry, the electric field is radial, and its magnitude is constant over the Gaussian surface. So, the integral simplifies to:

E * 4πr² = Q_encl / ε₀

4. Solving for the Electric Field (E)

Substitute our expression for Q_encl and solve for E:

E * 4πr² = [ρ * (4/3)πR³] / ε₀

Now, we solve for E:

E = (ρ * R³) / (3ε₀ * r²)

This is the electric field outside the sphere. Notice that it decreases with the square of the distance r from the center of the sphere. This is the same behavior we'd expect from a point charge!

5. Determining the Direction

Just like inside the sphere, the electric field points radially outward because the charge is positive.

Putting it All Together

We've now calculated the electric field both inside and outside the charged sphere. Let's summarize our results:

  • Inside the sphere (r < R): E = (ρ * r) / (3ε₀) (Linearly increasing with r)
  • Outside the sphere (r > R): E = (ρ * R³) / (3ε₀ * r²) (Decreasing with 1/r²)

To get the final answer, we need to plug in the values given in the problem:

  • R = 14.0 cm = 0.14 m
  • ρ = 2.262 x 10⁻³ C/m³
  • ε₀ = 8.854 x 10⁻¹² C²/N⋅m² (Permittivity of free space)

Let's calculate the electric field at the surface of the sphere (r = R) as an example:

E (at r = R) = (ρ * R) / (3ε₀) = (2.262 x 10⁻³ C/m³ * 0.14 m) / (3 * 8.854 x 10⁻¹² C²/N⋅m²) ≈ 1.19 x 10⁷ N/C

So, the electric field at the surface of the sphere is approximately 1.19 x 10⁷ N/C, pointing radially outward.

Conclusion

Calculating the electric field of a charged sphere might seem complicated, but by using Gauss's Law and understanding the symmetry of the problem, we can break it down into manageable steps. Remember the key concepts, choose your Gaussian surface wisely, and you'll be an electromagnetism pro in no time! This example is a fundamental building block for understanding more complex charge distributions and electric field scenarios. Keep practicing, and you'll master these concepts in no time!