Electric Field Inside And Outside A Uniformly Charged Sphere A Detailed Explanation

Hey everyone! Let's dive into an intriguing physics problem involving electric fields. This topic is super important for grasping electrostatics, and we're going to break it down step by step. Our main goal here is to explore how electric fields behave both inside and outside a uniformly charged non-conducting sphere. So, grab your thinking caps, and let’s get started!

The Problem: Electric Field Variations

So, here’s the puzzle we're tackling: Imagine we have a non-conducting sphere that’s uniformly charged. At a distance of 10 cm from the very center of this sphere, we measure an electric field, and we'll call its magnitude E. Now, the sphere itself has a radius of 5 cm. The big question is, what will the electric field be at a distance of 2.5 cm from the center? Sounds interesting, right? To solve this, we need to consider how electric fields behave differently inside and outside such a charged sphere. Let's dive into the concepts!

Key Concepts: Gauss's Law and Charge Distribution

To really nail this problem, we need to bring in a heavy hitter from the world of electromagnetism: Gauss's Law. This law is our best friend when dealing with symmetrical charge distributions, like our sphere. In simple terms, Gauss's Law helps us relate the electric field passing through a closed surface to the amount of charge enclosed within that surface. The magic formula that encapsulates this relationship is:

∮ E ⋅ dA = Qenc / ε0

Where:

• ∮ E ⋅ dA represents the electric flux—the measure of the electric field through a surface.
• Qenc is the total charge enclosed within the Gaussian surface.
• ε0 (epsilon naught) is the permittivity of free space, a constant value.

Now, why is this important? Well, Gauss's Law gives us a powerful tool to calculate the electric field if we can cleverly choose a Gaussian surface. For a sphere, a spherical Gaussian surface is the way to go because it maintains symmetry and simplifies our calculations.

But wait, there's more! We also need to think about how the charge is spread out in our sphere. Since it's a uniformly charged sphere, this means the charge is evenly distributed throughout its entire volume. This uniform charge distribution is crucial because it affects how we calculate the enclosed charge (Qenc) for different Gaussian surfaces—especially when we're looking at points inside the sphere.

To summarize, Gauss's Law provides the framework for our calculation, and the uniform charge distribution dictates how we apply the law in different regions of the sphere. Understanding these concepts is the key to unlocking the solution to our electric field puzzle!

Electric Field Outside the Sphere (r > R)

Let’s start by figuring out the electric field outside the charged sphere. This is the easier scenario, and it helps build our understanding before we tackle the inside. Imagine our point of interest is at a distance r from the center, where r is greater than the sphere's radius R. In our specific problem, we're first given the electric field at 10 cm, which is outside the 5 cm radius sphere.

To apply Gauss's Law, we draw an imaginary spherical surface—a Gaussian surface—centered on our charged sphere and passing through our point of interest. This Gaussian surface has a radius r. Because of the sphere's symmetry, the electric field will point radially outward (or inward, if the charge is negative) and will have the same magnitude at every point on our Gaussian surface. This makes our calculations much simpler!

Now, let's break down Gauss's Law for this scenario:

∮ E ⋅ dA = Qenc / ε0

Since the electric field E is constant and parallel to the area vector dA on our Gaussian surface, the integral simplifies to:

E ∮ dA = Qenc / ε0

The integral ∮ dA is just the surface area of our spherical Gaussian surface, which is 4πr². And Qenc, the charge enclosed, is simply the total charge Q on the sphere since our Gaussian surface completely encloses it. So, we have:

E (4πr²) = Q / ε0

Now, we can solve for the electric field E:

E = Q / (4πε0r²)

This is a familiar result! It tells us that the electric field outside the sphere behaves as if all the charge Q were concentrated at the center of the sphere. This is a crucial point: the electric field outside a uniformly charged sphere decreases with the square of the distance from the center. This inverse square relationship is a hallmark of many force laws in physics, like gravity and the electrostatic force.

So, we've figured out the electric field outside the sphere. We know it depends on the total charge, the distance from the center, and some constants. This gives us a foundation for understanding the more interesting case: what happens inside the sphere?

Electric Field Inside the Sphere (r < R)

Alright, guys, now for the juicy part: what happens to the electric field inside the uniformly charged sphere? This is where things get a little more interesting because not all the charge contributes to the electric field at our point of interest. Remember, we're looking at a point at a distance r from the center, where r is less than the sphere’s radius R. In our problem, this corresponds to the 2.5 cm distance, which is inside the 5 cm sphere.

We still use Gauss's Law, but the crucial difference is how we calculate the enclosed charge (Qenc). Let’s imagine our Gaussian surface again – a sphere of radius r centered on the charged sphere. But this time, our Gaussian surface is inside the charged sphere. So, it only encloses a portion of the total charge. This is key!

Since the charge is uniformly distributed, the amount of charge enclosed by our Gaussian surface is proportional to the volume it encloses. Let's say the total charge on the sphere is Q, and the sphere has a radius R. The charge density (charge per unit volume), which we'll call ρ (rho), is:

ρ = Q / (4/3πR³)

Now, the volume enclosed by our Gaussian surface (with radius r) is 4/3πr³. So, the charge enclosed (Qenc) is:

Qenc = ρ * (4/3πr³) = (Q / (4/3πR³)) * (4/3πr³) = Q * (r³/R³)

Notice that Qenc is not simply Q anymore! It's a fraction of Q that depends on the ratio of the cube of the distance r to the cube of the sphere’s radius R. This is a crucial result.

Now we can plug this into Gauss's Law:

E (4πr²) = Qenc / ε0 = (Q * (r³/R³)) / ε0

Solving for E, we get:

E = (Q * r) / (4πε0R³)

This is a fascinating result! The electric field inside the uniformly charged sphere is directly proportional to the distance r from the center. This is very different from the outside, where it decreased with the square of the distance. Inside, as you move closer to the center, the electric field gets weaker, reaching zero at the very center.

Why does this happen? Think about it this way: at any point inside the sphere, the electric field is created by the charge closer to the center than you are. The charge farther away effectively cancels out its contribution due to the spherical symmetry. As you get closer to the center, there's less and less charge