Question

A spherical cloud of electric charge has a known charge density \(Q(\rho),\) where \(\rho\) is the spherical coordinate. Find the total charge in the interior of the cloud in the following cases. a. \(Q(\rho)=\frac{2 \times 10^{-4}}{\rho^{4}}, 1 \leq \rho<\infty\) b. \(Q(\rho)=\left(2 \times 10^{-4}\right) e^{-0.01 \rho^{3}}, 0 \leq \rho<\infty\)

Short answer

**Question**: Calculate the total charge in the interior of a spherical cloud for two different cases: when \(Q(\rho) = \frac{2 \times 10^{-4}}{\rho^{4}}\), and when \(Q(\rho) = \left(2 \times 10^{-4}\right) e^{-0.01 \rho^{3}}\). **Answer**: For the first case, the total charge in the interior of the cloud is \(Q_a = 8 \pi \times 10^{-4}\) C. For the second case, the total charge is given by the expression \(Q_b = 8 \pi \times 10^{-4} \int_{0}^{\infty} e^{-0.01 \rho^{3}} \rho^2 d\rho\) C.

Step-by-step solution

Calculate infinitesimal charge

To find the infinitesimal charge \(dQ\) for each volume element \(dV\), use the charge density expression \(Q(\rho)\). For the spherical coordinate system, \(dV = 4 \pi \rho^2 d\rho\), so we have: \(dQ = Q(\rho) dV = Q(\rho) 4 \pi \rho^2 d\rho\)

Integrate for total charge in case a

For case a, we have \(Q(\rho) = \frac{2 \times 10^{-4}}{\rho^{4}}\). Plugging this expression into the formula for \(dQ\), we get: \(dQ_a = \frac{2 \times 10^{-4}}{\rho^{4}} 4 \pi \rho^2 d\rho\) Now, integrate \(dQ_a\) with respect to \(ρ\) from the given range \(1 \leq \rho < \infty\): \(Q_a = \int_{1}^{\infty} \frac{2 \times 10^{-4}}{\rho^{4}} 4 \pi \rho^2 d\rho = 8 \pi \times 10^{-4} \int_{1}^{\infty} \frac{1}{\rho^2} d\rho\) Integrate to find the total charge in case a: \(Q_a = 8 \pi \times 10^{-4} \left[-\frac{1}{\rho}\right]_{1}^{\infty} = 8 \pi \times 10^{-4}(0 + 1) = 8 \pi \times 10^{-4} C\)

Integrate for total charge in case b

For case b, we have \(Q(\rho) = \left(2 \times 10^{-4}\right) e^{-0.01 \rho^{3}}\). Plugging this expression into the formula for \(dQ\), we get: \(dQ_b = \left(2 \times 10^{-4}\right) e^{-0.01 \rho^{3}} 4 \pi \rho^2 d\rho\) Now, integrate \(dQ_b\) with respect to \(ρ\) from the given range \(0 \leq \rho < \infty\): \(Q_b = \int_{0}^{\infty} \left(2 \times 10^{-4}\right) e^{-0.01 \rho^{3}} 4 \pi \rho^2 d\rho = 8 \pi \times 10^{-4} \int_{0}^{\infty} e^{-0.01 \rho^{3}} \rho^2 d\rho\) This integral does not have a closed-form solution, so we will leave it as is to convey the expression for the total charge in case b: \(Q_b = 8 \pi \times 10^{-4} \int_{0}^{\infty} e^{-0.01 \rho^{3}} \rho^2 d\rho\) C

Key concepts

Charge Density

Charge density is a measure of the amount of electric charge per unit volume within a material or region of space. It is usually denoted by \( \rho(\mathbf{r}) \) in a general coordinate system. In this exercise, the charge density \( Q(\rho) \) varies with the radial distance \( \rho \) in spherical coordinates. Understanding charge density is crucial, as it helps us determine how charge is distributed within a given volume.

For example, in the first case, the charge density is given by \( Q(\rho) = \frac{2 \times 10^{-4}}{\rho^{4}} \). This implies a decrease in density as you move further from the center, because \( \rho^{4} \) increases.

In the second case, the charge density is \( Q(\rho) = \left(2 \times 10^{-4}\right) e^{-0.01 \rho^{3}} \), which also suggests a decrease, moderated by an exponential decay factor. This applies more immediate importance to the inner regions of our spherical cloud.

Integration

Integration plays a vital role in finding the total charge within a volume when charge density is known. By integrating the infinitesimal charges over the entire volume of interest, we can calculate the total charge.

To integrate, we first express the infinitesimal charge \( dQ \) using the equation \( dQ = Q(\rho) \, dV \), where \( dV \) is the volume element. In the step-by-step solution provided, integration is done over the radial component \( \rho \), as the charge density depends only on \( \rho \).

Each integral bounds the spherical radius, adjusting according to the problem requirements—\( 1 \leq \rho < \infty \) for case (a) and \( 0 \leq \rho < \infty \) for case (b). These bounds define how the integration captures the entire cloud's volume.

Spherical Coordinates

Spherical coordinates are a three-dimensional coordinate system that extends circular coordinates by adding a radial distance. It is especially useful for problems with spherical symmetry, such as the charge distribution in a cloud.

The coordinates are presented as \( (\rho, \theta, \phi) \):

• \( \rho \) is the radius or distance from the origin.
• \( \theta \) is the azimuthal angle in the \( xy \)-plane.
• \( \phi \) is the polar angle from the positive \( z \)-axis.

For this exercise, the main focus is on the radial part \( \rho \), as the charge density depends solely on this component. The simplicity of this setup allows us to efficiently calculate the charge within these coordinates.

Volume Element

The volume element \( dV \) in spherical coordinates is crucial for integration. It represents the differential volume over which an integral is performed. In spherical coordinates, the volume element is expressed as \( dV = \rho^2 \sin \theta \, d\rho \, d\theta \, d\phi \).

However, when the density is only a function of \( \rho \), such as in the given exercise, the task simplifies to considering only the radial component. This makes it \( dV = 4 \pi \rho^2 d\rho \) after integrating out the angular components, as they contribute merely a constant factor \( 4\pi \).

Utilizing this expression dramatically reduces the complexity of the calculations as we focus on the distribution along the radius, giving us a straightforward pathway to obtain the total charge.