Question

Two uniformly charged insulating rods are bent in a semicircular shape with radius \(r=10.0 \mathrm{~cm}\). If they are positioned so that they form a circle but do not touch and if they have opposite charges of \(+1.00 \mu \mathrm{C}\) and \(-1.00 \mu \mathrm{C},\) find the magnitude and the direction of the electric field at the center of the composite circular charge configuration.

Short answer

Answer: The magnitude of the electric field at the center is \(5.693 \times 10^4\, \text{N/C}\), and it points upwards.

Step-by-step solution

Identify given parameters and constants

We are given: Radius of each semicircular rod (r): \(10.0\,\text{cm} = 0.1\,\text{m}\), Charge of rod 1 (Q1): \(+1.00\,\mu\text{C} = +1.00 \times 10^{-6}\,\text{C}\), Charge of rod 2 (Q2): \(-1.00\,\mu\text{C} = -1.00 \times 10^{-6}\,\text{C}\), Electric constant (k): \(8.9875 \times 10^9\, \text{N m}^2/\text{C}^2\).

Divide each rod into small segments

To find the electric field, we will divide each semicircular rod into small segments (\(dq\)). The length of each segment can be considered infinitesimally small. We will first find the electric field at the center due to each small segment and then integrate along the entire semicircular rod.

Calculate the electric field due to a small segment

Consider a small segment \(dq_i\) on rod 1 at an angle \(\theta\) from the vertical line. The electric field (\(dE_{1i}\)) due to this segment at the center can be determined as: \(dE_{1i} = \dfrac{k \, dq_i}{r^2}\) Since the segments on rod 1 and rod 2 have same electric field magnitudes but opposite directions, the net electric field at the center is twice the vertical component of the electric field due to one segment (\(dE_{1iy}\)). \(dE_{1iy} = dE_{1i} \sin{\theta}\)

Calculate the total electric field at the center

To find the total electric field at the center (E), we need to integrate the expression for \(dE_{1iy}\) along the entire semicircular rod 1. \(E = 2 \int_0^{\pi} dE_{1iy}\) We know that \(dq_i = \dfrac{Q_1}{\pi r} \, d\theta\), where \(\dfrac{Q_1}{\pi r}\) is the charge per unit length. So, \(E = 2 \int_0^{\pi} \dfrac{k \, Q_1 \sin{\theta}}{\pi r^3} d\theta\) \(E = \dfrac{2kQ_1}{\pi r^3} \int_0^{\pi} \sin{\theta} d\theta\)

Calculate the integral and find the electric field

To finish the calculation, we need to find the integral: \(\int_0^{\pi} \sin{\theta} d\theta = [-\cos{\theta}]_0^{\pi} = 2\) Now, substitute the values to find the electric field: \(E = \dfrac{2 \times 8.9875 \times 10^9 \times 1.00 \times 10^{-6}}{\pi \times (0.1)^3} \times 2\) \(E = 5.693 \times 10^4\, \text{N/C}\) The direction of the electric field points upwards, which is the direction of the electric field created by the positively charged semicircular rod (rod 1). Thus, the magnitude of the electric field at the center is \(5.693 \times 10^4\, \text{N/C}\), and it points upwards.