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). c) Sketch the graph of \(E(r)\) versus \(r\). Comment on the continuity or discontinuity of the electric field, and relate this to the surface charge distribution on the gold layer.

Short answer

Answer: For ra (outside the coated sphere), the electric field becomes negative and decreases with radial distance following the equation \(E(r) = \frac{-Q}{4\pi r^2 \varepsilon_0}\). The discontinuity at the boundary between the nonconducting sphere and the gold layer is due to the change in electric field direction caused by the surface charge density of the gold layer.

Step-by-step solution

Define Gauss's Law

Gauss's Law states that the electric flux through a closed surface is equal to the total charge enclosed within the surface, divided by the permittivity of free space. Mathematically, it is given by: \(\oint \textbf{E} \cdot d\textbf{A} = \frac{Q_\text{enc}}{\varepsilon_0}.\)

Find the electric field inside the sphere (\(r

To find the electric field inside the sphere, we'll construct a Gaussian surface in the form of a sphere with radius r (\(r<a\)) and centered at the same point as the given sphere. In this case, the symmetry ensures that the electric field is radial, uniform, and perpendicular to the Gaussian surface. First, we need to find the charge enclosed within the Gaussian sphere with our given radius r: Since the total charge in the nonconducting sphere is given to be \(+Q\), and the charge is uniformly distributed throughout its volume, we can determine the charge density \(\rho\): \(\rho = \frac{Q}{\frac{4}{3}\pi a^3}\). Now, we can find the charge enclosed within the Gaussian sphere by multiplying the charge density by the volume of the Gaussian sphere: \(Q_\text{enc}=\frac{Q}{\frac{4}{3}\pi a^3}\times \frac{4}{3}\pi r^3 = Q\frac{r^{3}}{a^{3}}\). Next, we'll apply Gauss's Law: \(\oint \textbf{E} \cdot d\textbf{A} = \frac{Q_\text{enc}}{\varepsilon_0}\). Since the electric field is perpendicular and uniform on the Gaussian surface, the dot product and the integral can be simplified as follows: \(E \int dA = \frac{Q\frac{r^{3}}{a^{3}}}{\varepsilon_0}\). \(E(4\pi r^2) = \frac{Q\frac{r^{3}}{a^{3}}}{\varepsilon_0}\). Now, we solve for the electric field \(E(r)\): \(E(r) = \frac{Qr}{4\pi a^{3}\varepsilon_0}\) for \(r<a\), inside the sphere.

Find the electric field outside the sphere (\(r>a\))

Now, we'll find the electric field for the region outside the sphere (coated with a thin conducting layer of gold). We know that the total charge enclosed by a Gaussian sphere with radius r(\(r>a\)) is equal to the net charge of the system, which is (\(Q-2Q = -Q\)). According to Gauss's Law, we have: \(\oint \textbf{E} \cdot d\textbf{A} = \frac{-Q}{\varepsilon_0}\). Using a similar process as before, we get: \(E \int dA = \frac{-Q}{\varepsilon_0}\). \(E(4\pi r^2) = \frac{-Q}{\varepsilon_0}\). Solving for the electric field \(E(r)\), we get: \(E(r) = \frac{-Q}{4\pi r^2 \varepsilon_0}\) for \(r>a\), outside the coated sphere.

Sketch electric field E(r) versus r and discuss continuity

When plotting the electric field \(E(r)\) versus \(r\), you'll notice that the electric field increases linearly inside the sphere (\(ra\). This discontinuity occurs because the gold layer has a charge on its surface, meaning the electric field is changing direction at the boundary between the nonconducting sphere and the gold layer. This is consistent with the fact that the change in electric field across the surface of a conductor is related to the surface charge density.

Key concepts

Electric Field

Imagine a field surrounding every charged object, where the object’s influence can be felt -- this is the electric field. It’s an invisible force that can exert power on other charged particles or objects. In the exercise involving a nonconducting sphere with a uniform charge distribution and an outer conductive layer, understanding the electric field, denoted by the symbol E, is crucial.

The electric field resulting from a point charge can be represented by the equation E = k_e \( \frac{q}{r^2} \), where k_e is Coulomb's constant, q is the point charge, and r is the distance from the charge. However, for a uniformly charged sphere, we dive into a more intriguing scenario. Inside the sphere, the electric field increases linearly with distance from the center because only the charge enclosed within a smaller, imaginary sphere (radius r, which is less than the actual sphere’s radius a) contributes to the field at that point. Outside the charged sphere, the sphere acts as a point charge, and the electric field diminishes with the square of the distance, again.

This principle is neatly illustrated in the exercise, displaying the transition from a linearly increasing electric field within the nonconducting sphere to a decreasing field outside of it, influenced by the charge on the gold layer.

Charge Distribution

When we're casting a spotlight on charge distribution, envision how charge is 'spread out' in space, much like how stars might be scattered across the night sky. The way these charges are arranged, whether evenly throughout a volume, on a surface, or along a line, influences the electric field we’ve just discussed. In our exercise, there are two types of charge distribution to consider: the uniform volume distribution of charge within the nonconducting sphere, and the surface distribution on the gold layer.

Charge distribution is key because it shapes the electric field. In the exercise, the uniformly distributed charge inside the sphere creates an electric field that points outward and grows stronger as you move towards the sphere's surface. In contrast, the surface charge on the conductive layer results in an electric field that is constant over the surface and discontinuous at the boundary. This attribute leads us to interesting phenomena like the one described in the exercise, where the electric field experiences a sudden change – a discontinuity when transitioning from inside the sphere to the outside.

Electrostatics

The broader field encompassing these concepts is electrostatics, which delves into the forces, fields, and potentials arising from stationary electric charges. Electrostatics includes the study of how charges interact with each other and how they shape the spaces around them. One of the fundamental principles of electrostatics is Gauss's Law, which the exercise uses to derive the electric fields inside and outside the sphere.

Gauss's Law ties the electric field and charge distribution together into a neat bow. It states that the electric flux out of any closed surface is proportional to the charge enclosed by that surface. This law can be a great simplifier when dealing with symmetrical charge distributions, as it allows us to ignore everything except the charge that directly influences the region of interest.

In the given exercise, Gauss's Law helps to reveal the behavior of the electric field around the charged sphere coated with gold. It's a fantastic demonstration of the concepts at the heart of electrostatics, showing the inherent relationship between the electric field, the enclosed charge, and the resulting phenomena that these static electric configurations exhibit.