Question

Repeat Example \(22.3,\) assuming that the charge distribution is \(-\lambda\) for \(-a

Short answer

Based on the given information about the charge distribution and the position of point P, we derived an expression for the net electric field at that point by calculating the electric field produced by each charged segment and summing them up. The final expression for the net electric field requires integrating over the charge distribution and depends on the dimensionless ratio R/a and the angle θ. The net electric field at point P(R, θ) is measured in terms of the unit [λ/(4πε₀a)]. Due to the nature of the integral involved, it typically needs to be evaluated numerically.

Step-by-step solution

Define the problem and variables

We are asked to find the electric field at a given point \(P(R, \theta)\). Let \(d\vec{E}\) represent the electric field produced by a small charge element \(dq\) on the positively charged and negatively charged segments. Then, our goal is to find the net electric field \(E_{net}(R, \theta)\) by integrating \(d\vec{E}\) over the charged segments.

Set up the integral for the electric field

We can find the electric field by integrating \(d\vec{E}\) over each charged segment. First, we look at the negatively charged segment. We can use the following expression for the electric field created by a point charge \(dq\): \(d\vec{E} = \frac{1}{4\pi \epsilon_0} \frac{dq}{r^2} \hat{r}\), where \(r\) is the distance from the point charge to the point of interest and \(\hat{r}\) is the unit vector in the direction from the point charge to point P. By symmetry, we only need to consider the x-component of the electric field since the y-components will cancel out due to the charge distribution. The integral expression for the electric field created by the negatively charged segment in the x-direction can be expressed as: \(E_{1x} = -\int_{-a}^0 \frac{\lambda dx}{4\pi \epsilon_0}\frac{(R-r\cos\theta)}{[(R-r\cos\theta)^2+(r\sin\theta)^2]^{3/2}} \). Now, we look at the positively charged segment. As before, we only need to consider the x-component of the electric field as the y-components will cancel out due to the charge distribution. The integral expression for the electric field created by the positively charged segment in the x-direction can be expressed as: \(E_{2x} = -\int_{0}^a \frac{\lambda dx}{4\pi \epsilon_0}\frac{(R-r\cos\theta)}{[(R-r\cos\theta)^2+(r\sin\theta)^2]^{3/2}} \).

Find the net electric field

To determine the net electric field in the x-direction, we can add the electric fields created by both charged segments: \(E_{net} = E_{1x} + E_{2x} = -\int_{-a}^a \frac{\lambda dx}{4\pi \epsilon_0}\frac{(R-r\cos\theta)}{[(R-r\cos\theta)^2+(r\sin\theta)^2]^{3/2}} \).

Perform integration

We can now carry out the integration to find the net electric field at point \(P(R, \theta)\). The integral is not solvable in closed-form and usually needs to be evaluated numerically. However, it is important to note that the final answer depends only on the dimensionless ratio \(R/a\) and the unit of the electric field will be in terms of \(\left[\frac{\lambda}{4 \pi \epsilon_0 a}\right]\). Thus, the net electric field at point \(P(R, \theta)\) can be obtained by solving the integral: \(E_{net} = -\int_{-1}^1 \frac{d\xi}{\left[\left(\frac{R}{a}-\xi\cos\theta \right)^2+\xi^2\sin^2\theta \right]^{3/2}}\), where \(\xi = \frac{r}{a}\) and the net electric field is measured in terms of the unit \(\left[\frac{\lambda}{4\pi \epsilon_0 a}\right]\).