Question

A solid sphere of radius \(R\) has a nonuniform charge distribution \(\rho=A r^{2},\) where \(A\) is a constant. Determine the total charge, \(Q\), within the volume of the sphere.

Short answer

Question: Determine the total charge inside a solid sphere if the charge distribution function is given by \(\rho = A r^2\). Answer: The total charge inside the sphere is \(Q = \frac{4\pi A R^5}{5}\).

Step-by-step solution

Identify the method of integration

In this problem, we will use the spherical coordinate system to perform the integration, because it naturally fits the shape of the sphere.

Set up the volume element in spherical coordinates

In spherical coordinates, the volume element is given by \(dV = r^2 \sin \theta dr d\theta d\phi\). We will use this to convert the charge distribution into a function we can integrate.

Write down the limits of integration for each variable

Since we have to integrate over the entire volume of the sphere, the limits of integration for the variables are as follows: - For the radial distance (\(r\)): \(0\) to \(R\) - For the polar angle (\(\theta\)): \(0\) to \(\pi\) - For the azimuthal angle (\(\phi\)): \(0\) to \(2\pi\)

Set up the triple integral for total charge

Using the spherical coordinate system, the charge density function, and the limits of integration, we can set up the triple integral for the total charge as: $$Q = \int_{0}^{R} \int_{0}^{\pi} \int_{0}^{2\pi} \rho \ dV = \int_{0}^{R} \int_{0}^{\pi} \int_{0}^{2\pi} A r^2 r^2 \sin \theta \ dr d\theta d\phi$$

Solve the triple integral

Now, we can solve the triple integral step by step: First, integrate with respect to \(\phi\): $$Q = \int_{0}^{R} \int_{0}^{\pi} A r^4 \sin \theta \left[\int_{0}^{2\pi} d\phi\right] dr d\theta = \int_{0}^{R} \int_{0}^{\pi} A r^4 \sin \theta (2\pi) dr d\theta$$ Next, integrate with respect to \(\theta\): $$Q = \int_{0}^{R} A r^4 (2\pi) \left[\int_{0}^{\pi} \sin \theta d\theta\right] dr = \int_{0}^{R} A r^4 (2\pi) (-\cos \theta \Big|_{0}^{\pi}) dr$$ $$Q = \int_{0}^{R} A r^4 (2\pi) (2) dr$$ Finally, integrate with respect to \(r\): $$Q = A (4\pi) \left[\int_{0}^{R} r^4 dr\right] = A (4\pi) \left[\frac{r^5}{5}\Big|_{0}^{R}\right]$$

Calculate the total charge

Now, we substitute \(R\) into the antiderivative and simplify: $$Q = A (4\pi) \left[\frac{R^5}{5}\right] = \frac{4\pi A R^5}{5}$$ The total charge inside the sphere is \(Q = \frac{4\pi A R^5}{5}\).

Key concepts

Spherical Coordinates

Understanding spherical coordinates is key to solving problems involving three-dimensional symmetrical objects like spheres. Imagine grabbing a globe: you can pinpoint any location with just three values – how far from the center (the radius, r), how far north or south from the equator (the polar angle, θ), and how far around the sphere from a fixed longitude (the azimuthal angle, ϕ). In mathematics, these are called spherical coordinates, and they are essential when dealing with problems where symmetry makes Cartesian coordinates cumbersome.

In the case of a charge distribution inside a sphere, spherical coordinates allow us to express the problem in a way that complements the object's shape, leading to more straightforward integration. When we use spherical coordinates, we transform our view from flat grids to a system of nested spheres, cones, and circles, which perfectly matches the geometry of the problem at hand.

Triple Integral

When dealing with volumes in three-dimensional space, the triple integral comes into play. It extends the concept of single and double integrals into a realm where we are adding up infinitely small pieces of a three-dimensional object to determine a cumulative effect – in our case, the total charge. The triple integral will account for all dimensions: radial, polar, and azimuthal.

Understanding the Layers of Integration

It's like peeling an onion – we integrate with respect to one variable at a time, holding the others constant. Each integration brings us one layer closer to the solution. We start with ϕ, the outermost layer, which circles around the sphere. Then we move to θ, which goes up and down from pole to pole. Finally, we deal with r, which moves from the very center of the sphere out to its surface. This step-by-step approach simplifies the complex problem of finding the total charge in a sphere with a nonuniform charge distribution.

Volume Element

The volume element in spherical coordinates is the 'building block' we use to fill up the space inside the sphere. It's represented by dV and in spherical coordinates, it's given by dV = r² sin(θ) dr dθ dϕ. Imagine a tiny, infinitesimally small parcel of volume that's shaped somewhat like a wedge of an orange. It has a thickness of dr, a height determined by r dθ, and a width determined by the arc length r sin(θ) dϕ.

When we use the volume element in integration, we're adding up countless of these tiny parcels, fitting them together to fill the interior of the sphere without leaving any gaps or overlaps. And because the charge distribution might be different at every point inside the sphere, we multiply this volume element by the charge density function to determine the charge in each infinitesimal volume before integrating.

Charge Density Function

The charge density function, ρ, tells us how densely packed the charge is at any point within a given volume. For our sphere with a nonuniform charge distribution, it's not a constant value but varies depending on the distance from the center, as given by the function ρ = Ar². The A in the function is a constant that represents how the charge density changes with r. As r increases, the function suggests that the charge density gets more intense – it's denser further out from the center.

The beauty of modeling charge density with functions like this is that it allows us to account for complex, real-world scenarios where charge may not be evenly distributed. By integrating this function over the volume of the sphere, using our three spherical coordinates, we capture the total charge without missing any subtleties of the distribution.