Problem 33 - University Physics with Modern Physics

Question

A thin glass rod is bent into a semicircle of radius \(R\). A charge \(+Q\) is uniformly distributed along the upper half, and a charge \(-Q\) is uniformly distributed along the lower half as shown in the figure. Find the magnitude and direction of the electric field \(\vec{E}\) (in component form) at point \(P\), the center of the semicircle.

Step-by-Step Solution

Answer

Answer: The electric field at the center of the semicircle is \(\vec{E} = -\dfrac{2kQ}{R}\hat{j}\), where \(k\) is the Coulomb constant, \(Q\) is the total charge on each half of the semicircle, and \(R\) is the radius of the semicircle.

Step 1: Define a small charge element on the semicircle

Let's define a small charge element \(dq\) at an angle \(\theta\) on the semicircle. For the positive charges on the upper half, the angle \(\theta\) goes from \(0\) to \(\pi\) radians. For the negative charges on the lower half, the angle \(\theta\) goes from \(\pi\) to \(2\pi\) radians.

Step 2: Determine the electric field contribution due to a small charge element

The electric field contribution from a small charge element \(dq\) is given by \(d\vec{E} = k \dfrac{dq}{r^2}\hat{r}\), where \(k\) is the Coulomb constant, \(r = R\) (distance from charge element \(dq\) to point \(P\)), and \(\hat{r}\) is the unit vector pointing from \(dq\) to \(P\).

Step 3: Determine the components of the electric field due to the small charge element

To determine the \(X\) and \(Y\) components of the electric field, let's express \(d\vec{E}\) in component form: \(d\vec{E}_x = d\vec{E}\cos\theta = k \dfrac{dq}{R^2}\cos\theta\hat{i}\) \(d\vec{E}_y = d\vec{E}\sin\theta = k \dfrac{dq}{R^2}\sin\theta\hat{j}\)

Step 4: Integrate the electric field components over the entire semicircle

Now we need to integrate these components over the entire semicircle. We begin with the positive charges on the upper half \((0 \leq \theta \leq \pi)\): \(\vec{E}_{+x} = \int_{0}^{\pi} k \dfrac{dq}{R^2}\cos\theta\hat{i}\) \(\vec{E}_{+y} = \int_{0}^{\pi} k \dfrac{dq}{R^2}\sin\theta\hat{j}\) Then, we perform the same integration for the negative charges on the lower half \((\pi \leq \theta \leq 2\pi)\): \(\vec{E}_{-x} = \int_{\pi}^{2\pi} k \dfrac{dq}{R^2}\cos\theta\hat{i}\) \(\vec{E}_{-y} = \int_{\pi}^{2\pi} k \dfrac{dq}{R^2}\sin\theta\hat{j}\)

Step 5: Add the contributions from the positive and negative charges

To find the total electric field at point \(P\), we add the contributions from the positive and negative charges: \(\vec{E}_{x} = \vec{E}_{+x} + \vec{E}_{-x}\) \(\vec{E}_{y} = \vec{E}_{+y} + \vec{E}_{-y}\)

Step 6: Find the final expression for the electric field

After completing the integration, we find that the electric field components are: \(\vec{E}_{x} = 0\hat{i}\) \(\vec{E}_{y} = -\dfrac{2kQ}{R}\hat{j}\) Thus, the total electric field at point \(P\) is: \(\vec{E} = \vec{E}_{x} + \vec{E}_{y} = -\dfrac{2kQ}{R}\hat{j}\)