Problem 97 - University Physics with Modern Physics

Question

CALC Negative charge \(-Q\) is distributed uniformly around a quarter-circle of radius \(a\) that lies in the first quadrant, with the center of curvature at the origin. Find the \(x\) - and \(y\) -components of the net electric field at the origin.

Step-by-Step Solution

Answer

Both the x and y components of the electric field are \(\frac{2k_e Q}{\pi a^2}\), directed in the negative directions.

Step 1: Understanding the Setup

We have a quarter-circle in the first quadrant, spanning from the positive x-axis to the positive y-axis, with its center of curvature at the origin. The quarter-circle has a radius of \(a\) and carries a total negative charge \(-Q\). We need to find the electric field components at the origin.

Step 2: Infinitesimal Charge Element

Consider a small charge element \(dq\) on the quarter-circle. Since charge \(-Q\) is distributed uniformly, the linear charge density \(\lambda\) is \(\lambda = \frac{-Q}{\frac{\pi a}{2}} = \frac{-2Q}{\pi a}\). The charge element \(dq\) over an infinitesimal arc length \(ds = a d\theta\) is \(dq = \lambda \, ds = \frac{-2Q}{\pi a} \, a d\theta\).

Step 3: Determine Electric Field Contribution

The contribution to the electric field \(d\mathbf{E}\) at the origin from \(dq\) is given by Coulomb's law: \(dE = \frac{k_e \, |dq|}{a^2}\), where \(k_e\) is Coulomb's constant. The direction of \(d\mathbf{E}\) at the origin from \(dq\) at angle \(\theta\) is radially inward, forming an angle \(\theta\) with the negative x-axis.

Step 4: Components of the Electric Field

Decompose \(d\mathbf{E}\) into its components: \(dE_x = dE \cos\theta\) and \(dE_y = dE \sin\theta\). Substitute \(dE = \frac{k_e \, 2Q \, d\theta}{\pi a}\) to get the components: \(dE_x = \frac{k_e \, 2Q}{\pi a^2} \, \cos\theta \, d\theta\) and \(dE_y = \frac{k_e \, 2Q}{\pi a^2} \, \sin\theta \, d\theta\).

Step 5: Integrate to Find Total Electric Field Components

Integrate \(dE_x\) from \(0\) to \(\frac{\pi}{2}\): \[E_x = \int_0^{\pi/2} \frac{k_e \, 2Q}{\pi a^2} \, \cos\theta \, d\theta = \frac{k_e \, 2Q}{\pi a^2} \left[ \sin\theta \right]_0^{\pi/2} = \frac{k_e \, 2Q}{\pi a^2} \cdot 1 = \frac{2k_e Q}{\pi a^2}\]Integrate \(dE_y\) from \(0\) to \(\frac{\pi}{2}\):\[E_y = \int_0^{\pi/2} \frac{k_e \, 2Q}{\pi a^2} \, \sin\theta \, d\theta = \frac{k_e \, 2Q}{\pi a^2} \left[ -\cos\theta \right]_0^{\pi/2} = \frac{k_e \, 2Q}{\pi a^2} \cdot 1 = \frac{2k_e Q}{\pi a^2}\]

Step 6: Conclusion

The x-component of the net electric field at the origin is \(E_x = \frac{2k_e Q}{\pi a^2}\) directed in the negative x-direction, and the y-component is \(E_y = \frac{2k_e Q}{\pi a^2}\) directed in the negative y-direction in accordance with the field direction inward towards the charge distribution.

Key Concepts

Coulomb's Law

Coulomb's Law is foundational in understanding electric fields. It describes how electric charges interact at a distance. Imagine electric charge as a tiny invisible thread tying two charged spheres, either pulling them together or pushing them apart. This interaction is what we calculate through Coulomb's Law.
For point charges, the force between them is directly proportional to the product of their charges and inversely proportional to the square of the distance between them. Mathematically, it is expressed as:\[F = k_e \frac{|q_1 q_2|}{r^2}\]where:

• \(F\) is the magnitude of the force between the charges
• \(k_e\) is Coulomb's constant \(\approx 8.99 \times 10^9 \text{Nm}^2/\text{C}^2\)
• \(q_1\) and \(q_2\) are the amounts of the charges
• \(r\) is the distance between the charges

In our problem, we have a distribution of charge along a quarter-circle rather than point charges. We apply Coulomb's Law to calculate the effect of each infinitesimally small piece of charge along the curve. This requires summing, or rather integrating, these small effects to find the total electric field at a specific point.

Charge Distribution

Charge distribution refers to how electric charges are spread across a given area or line, influencing the surrounding electric field. In uniform charge distribution, like the one in our quarter-circle problem, each small segment of the arc contributes equally to the electric field at a point.
The linear charge density \(\lambda\) describes how much charge resides along a line segment, calculated as the total charge \(-Q\) divided by the length of the distribution. For our quarter-circle:\[\lambda = \frac{-Q}{\frac{\pi a}{2}} = \frac{-2Q}{\pi a}\]This formula helps us define a small charge element \(dq\) at a point \(\theta\) on the arc:\[dq = \lambda \cdot ds = \frac{-2Q}{\pi a} \cdot a d\theta\]By understanding how charge is distributed, we can determine how it influences the electric field. In essence, the shape, size, and nature of the distribution provides the framework for tackling complex problems involving electric fields.

Vector Integration

Vector integration allows us to calculate total effects from distributed sources in multiple dimensions. In our scenario, small charge segments on the quarter-circle each create electric fields pointing towards the origin. We need to add these contributions to determine the total electric field.
The magnitude of the electric field contributed by each segment is determined using Coulomb's law. Its direction points radially inward, aligning with the radius of the circle. Thus, we break the little electric field vectors into components along the x and y axes:\[dE_x = dE \cos\theta, \quad dE_y = dE \sin\theta\]By integrating these components over all infinitesimal charge elements along the quarter-circle, from angle \(0\) to \(\pi/2\), we find the total electric field. The integrations are:\[E_x = \int_0^{\pi/2} \frac{k_e 2Q}{\pi a^2} \cos\theta \, d\theta = \frac{2k_e Q}{\pi a^2}\]\[E_y = \int_0^{\pi/2} \frac{k_e 2Q}{\pi a^2} \sin\theta \, d\theta = \frac{2k_e Q}{\pi a^2}\]Vector integration transforms a complex, distributed charge setup into manageable calculations, yielding the resultant electric field at a focal point.