Question

A charge per unit length \(+\lambda\) is uniformly distributed along the positive \(y\) -axis from \(y=0\) to \(y=+a\). A charge per unit length \(-\lambda\) is uniformly distributed along the negative \(y\) -axis from \(y=0\) to \(y=-a\). Write an expression for the electric field (magnitude and direction) at a point on the \(x\) -axis a distance \(x\) from the origin.

Short answer

Based on the provided step-by-step solution, the electric field (magnitude and direction) at a point on the x-axis at a distance x from the origin, due to a linear charge distribution along the positive and negative y-axis, is given by: \(E_{x} = \frac{\lambda}{2\pi\epsilon_0} \left[ \frac{x}{\sqrt{x^2 + a^2}} - 1 \right] \hat{i}\) where, \(E_{x}\) is the electric field in the x-direction, \(\lambda\) is the charge density, \(\epsilon_0\) is the vacuum permittivity, \(a\) is the extent of the charge distribution on both the positive and negative y-axis, and \(\hat{i}\) is the unit vector along the x-axis.

Step-by-step solution

Define an infinitesimal charge element

For the positive charges along the positive y-axis, let us consider an infinitesimal charge element \(\delta q_1 = \lambda\delta y_1\) at a distance \(y_1\) from the origin. Similarly, for the negative charges along the negative y-axis, let us consider an infinitesimal charge element \(\delta q_2 = -\lambda\delta y_2\) at a distance \(y_2\) from the origin.

Find the contribution of electric field due to each charge element

The electric field due to each charge element at the point (\(x\), 0) can be defined as follows: \(\delta \mathbf{E_1} = \frac{1}{4\pi\epsilon_0} \frac{\delta q_1}{r_1^2} \hat{r_1} = \frac{1}{4\pi\epsilon_0} \frac{\lambda\delta y_1}{r_1^2} \hat{r_1}\) \(\delta \mathbf{E_2} = \frac{1}{4\pi\epsilon_0} \frac{\delta q_2}{r_2^2} \hat{r_2} = \frac{1}{4\pi\epsilon_0} \frac{-\lambda\delta y_2}{r_2^2} \hat{r_2}\) Here, \(r_1 = \sqrt{x^2 + y_1^2}\) and \(r_2 = \sqrt{x^2 + y_2^2}\) are the distances of the charge elements to the point on the x-axis, and \(\hat{r_1}\) and \(\hat{r_2}\) are the unit vectors pointing from the charge elements towards the point on the x-axis.

Resolve electric field contributions in the x direction

Since we're only interested in the magnitude and direction of the electric field at the point on the x-axis, we need to resolve the electric field components in the x direction. \(\delta E_{1x} = \delta E_1 \cos\theta_1 = \frac{1}{4\pi\epsilon_0} \frac{\lambda\delta y_1 x}{(x^2 + y_1^2)^{3/2}}\) \(\delta E_{2x} = \delta E_2 \cos\theta_2 = \frac{1}{4\pi\epsilon_0} \frac{-\lambda\delta y_2 x}{(x^2 + y_2^2)^{3/2}}\) Here, \(\theta_1\) and \(\theta_2\) are the angles between the horizontal line and the vectors \(\mathbf{r_1}\) and \(\mathbf{r_2}\) respectively.

Integrate the x-components of the electric field contributions

To find the total electric field at the point on the x-axis, we need to integrate the x-components of the electric field contributions over their respective charge distributions. \(E_{x} = E_{1x} + E_{2x} = \frac{\lambda}{4\pi\epsilon_0} \int_{0}^{a} \frac{x dy_1}{(x^2 + y_1^2)^{3/2}} - \frac{\lambda}{4\pi\epsilon_0} \int_{0}^{-a} \frac{x dy_2}{(x^2 + y_2^2)^{3/2}}\) \(E_{x} = \frac{\lambda}{4\pi\epsilon_0} \left[ \int_{0}^{a} \frac{x dy_1}{(x^2 + y_1^2)^{3/2}} + \int_{-a}^{0} \frac{x dy_2}{(x^2 + y_2^2)^{3/2}} \right]\)

Evaluate the integrals

After evaluating the integrals, we get the total electric field at the point on the x-axis: \(E_{x} = \frac{2\lambda}{4\pi\epsilon_0} \left[ \frac{x}{\sqrt{x^2 + a^2}} - \frac{x}{\sqrt{x^2}} \right]\) \(E_{x} = \frac{\lambda}{2\pi\epsilon_0} \left[ \frac{x}{\sqrt{x^2 + a^2}} - 1 \right]\) The final expression for the electric field (magnitude and direction) at a point on the x-axis at a distance x from the origin is: \(E_{x} = \frac{\lambda}{2\pi\epsilon_0} \left[ \frac{x}{\sqrt{x^2 + a^2}} - 1 \right] \hat{i}\)

Key concepts

Charge Distribution

Charge distribution refers to the way electric charge is spread out in space. It can be continuous, with charge spread uniformly over a volume, over a line, or on a surface, or it can be discrete, consisting of individual point charges. In our exercise, we are dealing with a linear charge distribution, where charge per unit length \( \lambda \) is uniformly distributed along the y-axis in both positive and negative directions. This distribution is important because it determines the electric field produced by the charges.

It is key for students to visualize charge distribution as if it were spread like bread crumbs along a path—in this case, the y-axis. The uniformity implies that if you picked any two segments of equal length along this path, they would contain the same amount of charge. This concept is foundational for calculating electric fields around such distributions.

Electric Field Due to Charge Element

An electric field due to a charge element represents the effect a small segment of charge has on the space around it. Think of it like the ripple created by a pebble in a pond—every little bit contributes to the overall effect. In electrostatics, we calculate the electric field (\( \delta \mathbf{E} \) ) at a point due to an infinitesimal charge element ( \( \delta q \) ) using Coulomb's Law. For a point charge, the electric field is proportional to the charge and inversely proportional to the square of the distance from the charge to the point of interest.

When students tackle problems involving charge distributions, they should conceptualize each infinitesimal element of charge as a point charge and calculate its electric field contribution, just as in our example where the infinitesimal charges are spread along the y-axis.

Superposition Principle in Electrostatics

The superposition principle is a crucial concept in electrostatics, allowing us to calculate the net electric field created by multiple charges. It states that the total electric field caused by a group of charges is the vector sum of the fields generated by each charge individually. This means you can compute the electric field due to each charge separately and then add them up vectorially to get the overall field.

The principle greatly simplifies dealing with complex charge distributions. In our exercise, the superposition principle means adding up the electric fields due to each infinitesimally small segment of charged line on both the positive and negative y-axis to get the total electric field on the x-axis. Students should envision this process like combining the ripples from multiple pebbles thrown into a pond; every ripple combines with others to form the resulting pattern.

Coulomb's Law

Coulomb's Law is the cornerstone of electrostatics, as it quantifies the amount of force between two stationary, electrically charged particles. The law's formula \(F = k_e \frac{q_1q_2}{r^2}\) relates the force (F) to the product of the charges (\(q_1\) and \(q_2\)) and inversely to the square of the distance (r) between them. The proportionality constant (\(k_e\)) is called Coulomb's constant.

In context, Coulomb's Law provides the foundation for calculating the electric field due to a point charge, which is essential when dealing with charge elements in a distribution. As an educator, emphasizing this relation helps students grasp how electric fields arise from individual charges and how those fields influence other charges in the space.

Electric Field Integration

Electric field integration is a mathematical process used to find the total electric field due to a continuous charge distribution. Since any distribution can be imagined as an assembly of point charges, by integrating, we're effectively adding up the infinitesimally small contributions from each point charge to determine the overall field.

In our problem, integration allows us to consider each infinitesimal charge along the y-axis. The result is a precise expression of how the electric field varies with position on the x-axis. This step is vital for students to master, as it bridges the concept with the actual computation, transforming the theory of superposition and Coulomb's Law into a practical result for real-world problems.