Question

A solid nonconducting sphere of radius \(a\) has a total charge \(+Q\) uniformly distributed throughout its volume. The surface of the sphere is coated with a very thin (negligible thickness) conducting layer of gold. A total charge of \(-2 Q\) is placed on this conducting layer. Use Gauss's Law to do the following. a) Find the electric field \(E(r)\) for \(ra\) (outside the coated sphere, beyond the sphere and the gold layer).

Short answer

The electric field inside the sphere (r < a) can be expressed as \(E(r) = \frac{Qr}{4\pi \epsilon_0 a^3}\). The electric field outside the sphere (r > a) can be expressed as \(E(r) = -\frac{Q}{4\pi \epsilon_0 r^2}\).

Step-by-step solution

Consider a Gaussian surface inside the sphere

Take a spherical Gaussian surface with radius r inside the sphere (r<a). We will find the electric field at a point on this Gaussian surface.

Apply Gauss's Law to the Gaussian surface

Write Gauss's Law: \(\oint \vec{E} \cdot \vec{dA} = \frac{Q_{enc}}{\epsilon_0}\) Here, \(Q_{enc}\) is the charge enclosed by the Gaussian surface and \(\epsilon_0\) is the permittivity of free space.

Calculate the charge enclosed in the Gaussian surface

Since the charge is uniformly distributed throughout the sphere, we can write the total charge enclosed as: \(Q_{enc}=\rho \cdot V\) Here, V is the volume of the Gaussian surface, and \(\rho\) is the charge density given by \(\frac{Q}{\frac{4}{3}\pi a^3}\) (total charge divided by the volume of the sphere). \(Q_{enc} = \frac{Q}{\frac{4}{3}\pi a^3} \cdot \frac{4}{3}\pi r^3\) \(Q_{enc} = \frac{Qr^3}{a^3}\)

Finding the electric field inside the sphere

Now, substitute \(Q_{enc}\) in Gauss's Law: \(\oint \vec{E} \cdot \vec{dA} = \frac{Qr^3}{a^3\epsilon_0}\) Since the electric field is radially outward, it is parallel to the area vector at every point on the Gaussian surface. So, \(\oint \vec{E} \cdot \vec{dA} = E \oint dA\). By symmetry, the electric field is constant on the Gaussian surface. Therefore, we can write: \(E\oint dA = E(4\pi r^2)\) Now, equating the two expressions, we have: \(E(4\pi r^2) = \frac{Qr^3}{a^3\epsilon_0}\) Solving for E(r), we get: \(E(r) = \frac{Qr}{4\pi \epsilon_0 a^3}\), for r < a b) Finding the electric field E(r) for r>a (outside the coated sphere, beyond the sphere and the gold layer).

Consider a Gaussian surface outside the sphere

Consider a spherical Gaussian surface with radius r outside the sphere, where r>a.

Apply Gauss's Law to the Gaussian surface

Like before, write Gauss's Law: \(\oint \vec{E} \cdot \vec{dA} = \frac{Q_{enc}}{\epsilon_0}\)

Calculate the total charge enclosed by the Gaussian surface

This time, the Gaussian surface encloses both the charge \(+Q\) inside the nonconducting sphere and the charge \(-2Q\) on the gold layer. So, the total charge enclosed is: \(Q_{enc} = +Q - 2Q = -Q\)

Finding the electric field outside the sphere

Substitute \(Q_{enc}\) in Gauss's Law: \(\oint \vec{E} \cdot \vec{dA} = -\frac{Q}{\epsilon_0}\) Again, the electric field is radially outward, so \(\oint \vec{E} \cdot \vec{dA} = E \oint dA = E(4\pi r^2)\). Equating the two expressions, we get: \(E(4\pi r^2) = -\frac{Q}{\epsilon_0}\) Solving for E(r), we obtain: \(E(r) = -\frac{Q}{4\pi \epsilon_0 r^2}\), for r > a Hence, the electric field for ra is \(E(r) = -\frac{Q}{4\pi \epsilon_0 r^2}\).

Key concepts

Electric Field Inside a Sphere

Understanding the behavior of the electric field inside a sphere can be perplexing, but it's an essential part of electromagnetism. So let's dive into it. Imagine a solid nonconducting sphere that's evenly charged throughout its volume. According to Gauss's Law, the electric field within such a sphere isn’t just a random value; it's intimately connected to the amount of charge enclosed within an imaginary Gaussian surface.

Here's the fascinating part: Inside the sphere, as you move closer or further from the center, the electric field actually changes. This is contrary to what happens in a hollow sphere, where the electric field inside would be zero. Going through the calculations using Gauss's Law, we find that the electric field for a point at a distance 'r' from the center is proportional to 'r', meaning it increases linearly from the center to the surface. This linear relationship between electric field and the radius inside the sphere emphasizes the uniform nature of the charge distribution within the object.

Electric Field Outside a Sphere

When venturing outside the sphere, particularly one with an interesting charge makeup like our hypothetical gold-coated one, things get different. Here’s the twist: the charges inside and on the surface behave as if they were concentrated at the center for the purpose of calculating the electric field at points outside the sphere. This makes the situation similar to having a point charge with the sphere’s total charge.

Applying Gauss’s Law once again reveals that for distances 'r' greater than the sphere's radius, the electric field follows an inverse square law. It decreases as the square of the distance from the sphere's center increases. So, the farther you are from the sphere, the weaker the electric field becomes. This insight is pivotal in understanding the long-range effects of electrostatic forces and how they diminish with distance.

Charge Distribution

The concept of charge distribution is central to predicting the behavior of electric fields around charged objects. In our sphere example, the inner positive charge is uniformly dispersed, resulting in a neat, proportional relationship inside the object. However, once you coat the sphere with a thin layer of gold that carries a negative charge, you're adding a layer of complexity—literally.

Understanding how charges distribute themselves is key to mastering Gauss's Law. For instance, in conductors, charges reside on the surface, influencing the resultant electric field around them. On the contrary, the charges are spread throughout the volume of nonconducting or insulating materials. This dynamic is what dictates the behavior of the electric field both inside and outside the sphere and underscores the importance of the distribution of charges in electrostatics.

Permittivity of Free Space

The permittivity of free space, often symbolized as \(\epsilon_0\), is a fundamental constant in electromagnetism that describes how much resistance is encountered when forming an electric field in a vacuum. It essentially dictates the strength of the electric field that can be generated per unit of charge.

The value of the permittivity of free space comes into play when you apply Gauss's Law, serving as a mediator for how charges give rise to electric fields. This constant ensures that the formulas you use align with real-world behaviors observed in vacuum conditions. However, it is also used in media other than a vacuum to introduce the concept of relative permittivity or dielectric constant, which describes how a material modifies the electric field within it compared to a vacuum. Understanding \(\epsilon_0\) allows us to quantify the strength of electric fields produced by charges in any circumstance and is a cornerstone feature of electrostatics.