Question

There is a uniform charge distribution of \(\lambda=5.635 \cdot 10^{-8} \mathrm{C} / \mathrm{m}\) along a thin wire of length \(L=22.13 \mathrm{~cm} .\) The wire is then curved into a semicircle that is centered at the origin and has a radius of \(R=L / \pi .\) Find the magnitude of the electric field at the center of the semicircle.

Short answer

Answer: The magnitude of the electric field at the center of the semicircle is 4.76 x 10³ N/C.

Step-by-step solution

Determine the infinitesimally small charge segment

Consider a small arc length \(dL\) at an angle \(\theta\) from the center of the semicircle. The charge on this small segment would be given by \(dq = \lambda dL\).

Define the electric field to the center due to the small charge segment

The electric field due to the infinitesimal charge \(dq\) at the center of the semicircle is given by: \(dE = \frac{k dq}{R^2}\) where k is the electric constant.

Determine the electric field components due to the segment

The infinitesimal electric field can be broken down into its x and y-components, which are given by: \(dE_x = dE \cos{\theta}\), and \(dE_y = dE \sin{\theta}\).

Integrate the electric field components over the semicircle

We integrate the x and y-components of the electric field over the semicircle (\(0\) to \(\pi\)) to find the total electric field components at the center of the semicircle: \(E_x = \int_{0}^{\pi} dE_x = \int_{0}^{\pi} \frac{k dq}{R^2} \cos{\theta}\) \(E_y = \int_{0}^{\pi} dE_y = \int_{0}^{\pi} \frac{k dq}{R^2} \sin{\theta}\)

Substitute dq and express dL in terms of dθ

Recall that \(dq = \lambda dL\). We can express dL in terms of dθ using the relation \(dL = R d\theta\). Substituting these into the integrals, we get: \(E_x = \int_{0}^{\pi} \frac{k \lambda R d\theta}{R^2} \cos{\theta}\) \(E_y = \int_{0}^{\pi} \frac{k \lambda R d\theta}{R^2} \sin{\theta}\)

Evaluate the integrals and simplify

Evaluate the integrals and simplify the expressions: \(E_x = \frac{k \lambda}{R} \int_{0}^{\pi} \cos{\theta} d\theta = 0\) \(E_y = \frac{k \lambda}{R} \int_{0}^{\pi} \sin{\theta} d\theta = 2 \frac{k \lambda}{R}\)

Find the magnitude of the total electric field

Since the x-component of the electric field is zero, the magnitude of the electric field is equal to the y-component: \(E = E_y = 2 \frac{k \lambda}{R}\)

Substitute given values and find the electric field magnitude

Substitute the given values of \(\lambda = 5.635 \cdot 10^{-8} \mathrm{C} / \mathrm{m}\), and \(R = \frac{L}{\pi} = \frac{22.13 \mathrm{cm}}{\pi}\) into the equation and find the magnitude of the electric field: \(E = 2 \frac{(8.99 \times 10^9 Nm^2/C^2)(5.635 \times 10^{-8} C/m)}{\frac{22.13 \times 10^{-2} m}{\pi}}\) Solving this equation, we get: \(E = 4.76 \times 10^3 \mathrm{N/C}\) Therefore, the magnitude of the electric field at the center of the semicircle is \(4.76 \times 10^3 \mathrm{N/C}\).

Key concepts

Uniform Charge Distribution

Understanding the concept of a uniform charge distribution is crucial when studying electric fields. It refers to a scenario where the charge per unit length, surface, or volume remains constant over an object. In the given exercise, the uniform charge distribution is along the length of a wire.

This is a simplifying assumption that allows us to use linear charge density, denoted as \(\lambda\), which in this case equals \(5.635 \cdot 10^{-8} \mathrm{C} / \mathrm{m}\). It means that every meter of the wire carries the same amount of charge. As the wire forms a semicircle, this uniform distribution helps simplify the complex geometry into small segments where precise calculations can be done.

When the wire is curved into a semicircle, each infinitesimal segment still carries an amount of charge proportional to its length, making the use of integration possible to calculate the total electric field. It's important to note that in reality, achieving an absolutely uniform distribution can be challenging, but it is a helpful theoretical approximation.

Electric Field Components

In the study of electric fields, the idea of breaking down vectors into components is frequently used. The components of an electric field show us how it behaves along the x and y axis in a Cartesian plane. When a charge distribution is symmetric, like the semicircle in the exercise, the symmetry can be exploited to easily resolve the electric field components.

The infinitesimal electric field, \(dE\), created by a small charge segment, can be broken down into \(dE_x\) and \(dE_y\), its horizontal and vertical components, respectively. This helps us understand that the electric field, a vector quantity, can influence charged particles in both directions, and each of these influences can be analyzed separately.

For the semicircle, since it's symmetrical across the y-axis, the x-components of the electric field produced by opposite segments cancel each other out. As a result, only the y-components contribute to the net electric field at the center. This illustrates how symmetry can simplify the problem and focus our efforts on relevant calculations.

Integration in Electric Fields

Integration is a mathematical technique used to sum up infinitesimally small quantities to find a total. In the context of electric fields created by a distributed charge, integration plays a pivotal role. It allows us to calculate the total electric field generated by adding up all the contributions from each infinitesimal charge segment.

In our semicircle example, after breaking down the electric field components, we integrate \(dE_x\) and \(dE_y\) over the angle \(\theta\) that runs from 0 to \(\pi\) (for the semicircle). This is because each small charge segment corresponds to a small angle, and together, the entire semicircle spans \(\pi\) radians. The integral of the x-components ends up being zero due to symmetry, and we only integrate the y-components from 0 to \(\pi\) to find the net electric field at the center.

The outcome of this process is a mathematical expression for the electric field that can then be used to evaluate its magnitude at a specific point by substituting the relevant values. In essence, integration enables us to transition from infinitely small increments to a complete picture of an electric field produced by a distributed charge.