Problem 35 - University Physics with Modern Physics

Question

A uniformly charged rod of length \(L\) with total charge \(Q\) lies along the \(y\) -axis, from \(y=0\) to \(y=L\). Find an expression for the electric field at the point \((d, 0)\) (that is, the point at \(x=d\) on the \(x\) -axis).

Step-by-Step Solution

Answer

The electric field at point (d, 0) due to the uniformly charged rod is given by: $$\vec{E} = \frac{\lambda d}{4\pi\epsilon_0} \bigg(\frac{1}{d} - \frac{1}{\sqrt{d^2 + L^2}}\bigg) \hat{x}$$ where $\lambda$ is the charge density, $L$ is the length of the rod, and $\epsilon_0$ is the vacuum permittivity.

Step 1: Begin by calculating the charge density

Since the rod is uniformly charged, we can find the charge density using the following formula: $$\lambda = \frac{Q}{L}$$ where \(\lambda\) is the charge density, \(Q\) is the total charge, and \(L\) is the length of the rod.

Step 2: Find the electric field due to \(\delta q\)

Now we'll consider a small element \(\delta y\) of the rod at distance \(y\) from the reference point and having a charge \(\delta q = \lambda \delta y\). The electric field due to this small element \(\delta q\) at point \((d, 0)\) can be calculated using Coulomb's law: $$\delta \vec{E} = \frac{1}{4\pi\epsilon_0} \frac{\delta q}{r^2} \hat{r}$$ where \(\epsilon_0\) is the vacuum permittivity, \(r\) is the distance between point \((d, 0)\) and the charge element \(\delta q\), and \(\hat{r}\) is the unit vector from the charge element to the point.

Step 3: Calculate the distance and the unit vector

To calculate \(r\) and \(\hat{r}\), we first need to find the coordinates of the charge element: \((0, y)\). The distance between the point \((0, y)\) and point \((d, 0)\) is: $$r = \sqrt{d^2 + y^2}$$ And the unit vector \(\hat{r}\) can be found as: $$\hat{r} = \frac{\vec{r}}{r} = \frac{d\hat{x} + y\hat{y}}{\sqrt{d^2 + y^2}}$$

Step 4: Plug the values into the electric field expression

Substituting the values of \(r\) and \(\hat{r}\) in the expression for the electric field due to the charge element \(\delta q\), we get: $$\delta \vec{E} = \frac{1}{4\pi\epsilon_0} \frac{\lambda \delta y}{d^2 + y^2}(\frac{d\hat{x} + y\hat{y}}{\sqrt{d^2 + y^2}})$$

Step 5: Integrate the electric field expression along the rod

Now we'll integrate this electric field expression along the length of the rod to find the total electric field at point \((d, 0)\). We need to integrate individually for the components along the \(\hat{x}\) and \(\hat{y}\) directions: $$\vec{E} = \int_0^L \delta \vec{E} = \frac{\lambda}{4\pi\epsilon_0} \int_0^L \frac{d\hat{x} + y\hat{y}}{(d^2 + y^2)^{3/2}} \delta y$$ Separate the components and integrate: $$E_x = \frac{\lambda}{4\pi\epsilon_0} \int_0^L \frac{d}{(d^2 + y^2)^{3/2}} \delta y$$ $$E_y = \frac{\lambda}{4\pi\epsilon_0} \int_0^L \frac{y}{(d^2 + y^2)^{3/2}} \delta y$$

Step 6: Solve the integrals

Solve the integration for both components: $$E_x = \frac{\lambda d}{4\pi\epsilon_0} \bigg[-\frac{1}{\sqrt{d^2 + y^2}}\bigg]_0^L = \frac{\lambda d}{4\pi\epsilon_0} \bigg(\frac{1}{d} - \frac{1}{\sqrt{d^2 + L^2}}\bigg)$$ $$E_y = 0$$ (Symmetry along the \(y\)-axis)

Step 7: Write the final expression for the electric field

The electric field at point \((d, 0)\) can now be written as: $$\vec{E} = E_x\hat{x} + E_y\hat{y} = \frac{\lambda d}{4\pi\epsilon_0} \bigg(\frac{1}{d} - \frac{1}{\sqrt{d^2 + L^2}}\bigg) \hat{x}$$ This is the electric field due to the uniformly charged rod at point \((d, 0)\).

Key Concepts

Coulomb's Law

Coulomb's Law is a fundamental principle used to understand how electric charges interact. It states that the electric force between two point charges is directly proportional to the product of their charges and inversely proportional to the square of the distance between them. Here is what this means:

• When charges have the same sign (both positive or both negative), they repel each other.
• Oppositely charged objects attract each other.
• The formula for this interaction can be expressed as: \[ F = \frac{k \, |q_1 \, q_2|}{r^2} \]where \( F \) is the force, \( q_1 \) and \( q_2 \) are the amounts of charge, \( r \) is the distance between the charges, and \( k \) is Coulomb's constant \( 8.99 \times 10^9 \, \text{N m}^2/\text{C}^2 \).

This fundamental law guides the calculation of electric fields, which represent this force in a field of view. By understanding Coulomb's Law, students can correctly link the influence of charge and distance with the resulting electric field around them.

Uniform Charge Distribution

A uniform charge distribution means that the charge is spread evenly over the length, area, or volume of an object. This concept is applied frequently in problems involving electric fields. Here's what it entails:

• With a linear distribution (such as a rod), charge is evenly distributed along a line. For example, the charge density \( \lambda \) is given by \( \lambda = \frac{Q}{L} \), where \( Q \) is the total charge and \( L \) is the length of the rod.
• Understanding uniform charge distribution makes it easier to divide a complex problem into smaller, manageable sections (or elements) that are easier to analyze.

This concept is particularly helpful in simplifying calculations, as seen in solving the expression for the electric field due to a charged rod. Recognizing a system's symmetry can also streamline the calculation of its electric field, as symmetric distributions often lead to some components of the field canceling out.

Integration in Electromagnetism

Integration is a powerful mathematical tool used to calculate physical quantities that are distributed over a region, such as charge distributions for electric fields. In electromagnetism, this involves:

• Breaking down a continuous charge distribution into tiny segments and calculating the field for each segment using known formulas, like the one derived from Coulomb's Law.
• Summing the contributions of all these small segments using integration, which replaces the discreet sum of point charges with a continuous model.

For our exercise, integration is used to sum the electric field contributions along the length of a charged rod. We calculate the tiny electric field \( \delta\vec{E} \) due to each small charge element \( \delta q \) associated with a small segment \( \delta y \) of the rod. Then, we integrate these contributions to find the net electric field. In this case, it gives:

• \( E_x \), the electric field component along the \( x \)-axis, requires the integral of \( \frac{d}{(d^2 + y^2)^{3/2}} \).
• \( E_y \), the component along the \( y \)-axis, turns out to be zero due to symmetry, thus confirming that no vertical component affects the net field at point \( (d, 0) \).

Understanding integration in such contexts not only clarifies how individual elements contribute to a whole but also depicts how fields behave in complex configurations.