Problem 51 - Fundamentals of Physics

Question

A charged nonconducting rod, with a length of \(2.00 \mathrm{~m}\) and a cross- sectional area of \(4.00 \mathrm{~cm}^{2}\), lies along the positive side of an \(x\) axis with one end at the origin. The volume charge density \(\rho\) is charge per unit volume in coulombs per cubic meter. How many excess electrons are on the rod if \(\rho\) is (a) uniform, with a value of \(-4.00 \mu \mathrm{C} / \mathrm{m}^{3},\) and \((\mathrm{b})\) nonuniform, with a value given by \(\rho=b x^{2}\) where \(b=-2.00 \mu \mathrm{C} / \mathrm{m}^{5} ?\)

Step-by-Step Solution

Answer

(a) 2.00 x 10^10 electrons, (b) 1.33 x 10^10 electrons.

Step 1: Calculate the total charge for uniform density

For a uniform charge density, the total charge \( Q \) is given by \( Q = \rho \times V \), where \( V \) is the volume of the rod. The volume \( V \) is calculated as \( V = A \times L \), where \( A = 4.00 \times 10^{-4} \text{ m}^2 \) and \( L = 2.00 \text{ m} \). Thus, \( V = 4.00 \times 10^{-4} \times 2.00 = 8.00 \times 10^{-4} \text{ m}^3 \). With \( \rho = -4.00 \times 10^{-6} \text{ C/m}^3 \), the total charge is \( Q = -4.00 \times 10^{-6} \times 8.00 \times 10^{-4} = -3.20 \times 10^{-9} \text{ C} \).

Step 2: Convert charge to number of electrons (uniform density)

The charge of one electron is \( -1.60 \times 10^{-19} \text{ C} \). The number of electrons \( N \) is given by \( N = \frac{Q}{e} \), where \( e = -1.60 \times 10^{-19} \text{ C} \). Substitute \( Q = -3.20 \times 10^{-9} \text{ C} \) into the equation to get \( N = \frac{-3.20 \times 10^{-9}}{-1.60 \times 10^{-19}} = 2.00 \times 10^{10} \) electrons.

Step 3: Calculate the total charge for nonuniform density

For nonuniform density, integrate the charge density over the length of the rod: \( Q = \int_0^L \rho(x) \, A \, dx \). With \( \rho(x) = b x^2 \) and \( b = -2.00 \times 10^{-6} \text{ C/m}^5 \), compute \( Q = \int_0^{2.00} (-2.00 \times 10^{-6} x^2) (4.00 \times 10^{-4}) dx \). Evaluate this integral: \( Q = -8.00 \times 10^{-10} \int_0^{2.00} x^2 \, dx = -8.00 \times 10^{-10} \left[ \frac{x^3}{3} \right]_0^{2.00} = -8.00 \times 10^{-10} \left( \frac{8}{3} \right) = -2.13 \times 10^{-9} \text{ C} \).

Step 4: Convert charge to the number of electrons (nonuniform density)

Using the same method as in Step 2, compute the number of electrons. Substitute \( Q = -2.13 \times 10^{-9} \text{ C} \) into \( N = \frac{Q}{e} \) to get \( N = \frac{-2.13 \times 10^{-9}}{-1.60 \times 10^{-19}} = 1.33 \times 10^{10} \) electrons.

Key Concepts

Uniform Charge Distribution

In a uniform charge distribution, the charge is spread evenly over the volume of an object. Imagine the charge distribution as if a perfect grid was laid out, where every portion of the grid holds the same amount of charge. This results in a constant charge density across the entire object.

This uniformity simplifies the mathematical treatment of the charge since the charge density \( \rho \) remains consistent. In practice, we calculate the total charge \( Q \) by multiplying the uniform charge density by the volume of the object: \( Q = \rho \times V \). Here, \( V \) represents the total volume. For example, if the charge density for a rod is \(-4.00 \mu \text{C/m}^3\), and the rod’s volume is \(8.00 \times 10^{-4} \text{ m}^3\), the total charge \( Q \) is computed directly as \(-3.20 \times 10^{-9} \text{ C}\).

This approach is ideal for simple geometries and homogenous materials where the charge distribution doesn't vary with position. Uniform distribution makes calculations straightforward and predictable since there is no need to consider variations at different points of the object.

Nonuniform Charge Distribution

Nonuniform charge distribution occurs when charge density varies throughout an object. This variation can depend on specific parameters such as position or temperature. For nonuniform charge distributions, charges are scattered unevenly, and charge density might change at different locations.

In mathematical terms, the charge density is expressed as a function of position, for instance \( \rho(x) = b x^2 \), where the charge density varies with the square of the position \( x \). Here, calculations involve integrating the charge density function over the object's volume to find the total charge \( Q \).

Consider a rod with a nonuniform charge density \( ho(x) = -2.00 \mu \text{C/m}^5 \times x^2 \). To find the total charge, we integrate: \( Q = \int_0^{L} \rho(x) \, A \, dx \). For this rod, this process results in a total charge of \(-2.13 \times 10^{-9} \text{ C}\).
Handling nonuniform distributions can be complex, as each part of the object might have different charge values, but integration provides a powerful method to account for these variations.

Volume Charge Density

Volume charge density represents how much charge is contained per unit volume within an object. This concept is analogous to density in physics but applies to charge, which means it helps us understand how "thickly" the charge is packed within objects.

Symbolically, it is denoted by \( \rho \) and measured in coulombs per cubic meter (C/m³). For a uniform charge distribution, this value is constant throughout the object. In contrast, a nonuniform distribution features a charge density that changes with position, described by a mathematical function like \( \rho(x) = bx^2 \).

Knowing the volume charge density is crucial when calculating the total charge of an object or when predicting how the object will interact with electric fields. In practical terms, for a rod with a cross-sectional area of \(4.00 \text{ cm}^2\) and length of \(2.00 \text{ m}\), finding the volume charge density is a starting point to determine the total charge held by the rod.

Integration in Physics

Integration in physics is a powerful tool used to sum quantities over a continuous range. It is particularly useful for problems involving nonuniform distributions such as charge density, where simple multiplication isn't sufficient.

For a variable charge density like \( \rho(x) = b x^2 \), you can't determine the total charge by merely multiplying by volume. Instead, you need to integrate this function over the object's volume. The integral aggregates values of \( \rho \) from one end of the object to the other, giving an accurate representation of the total quantity, such as charge.
The integral for the charge over a rod from \( x = 0 \) to \( x = L \) would be expressed as \( Q = \int_0^L \rho(x) \, A \, dx \). This integral is solved step by step, taking into account the variable nature of \( \rho(x) \).
Integration thus bridges the gap between theoretical expressions and practical calculations, allowing for precise measurements and predictions in real-world scenarios involving complex distributions.