Let's break down the problem and solve for the **electric field at the origin O** due to the described charge distribution. --- ## **Problem Restatement- **Configuration - Two semi-infinite lines parallel to \(x\)-axis at \(y = +a\) and \(y = -a\), from \(x = \) to \(x \to +\infty\) - Connected at their left ends (\(x = \)) by a semicircular arc of radius \(a\), centered at \(O\) - **Charge density:** Uniform, \(\rho_L\) - **Find:** Electric field at \(O\). --- ## **Step 1: Semicircular Arc Contribution** For a uniformly charged arc of radius \(a\) subtending angle \(\pi\) (semicircle): - **Element:** \(dq = \rho_L a d\theta\), where \(\theta\) runs from \(\pi/2\) to \(3\pi/2\). - **Position:** On the arc, \( \vec{r} = a(\cos\theta\,\hat{i} + \sin\theta\,\hat{j}) \) - **Electric field due to \(dq\) at \(O\):** \[ d\vec{E} = \frac{1}{4\pi\epsilon_} \frac{dq}{a^2} (-\cos\theta\,\hat{i} - \sin\theta\,\hat{j}) \] (negative because direction is from element to \(O\)) Substitute \(dq\): \[ d\vec{E} = \frac{\rho_L}{4\pi\epsilon_ a} (-\cos\theta\,\hat{i} - \sin\theta\,\hat{j}) d\theta \] - **Total arc field:** \[ \vec{E}_{\text{arc}} = \frac{\rho_L}{4\pi\epsilon_ a} \int_{\pi/2}^{3\pi/2} (-\cos\theta\,\hat{i} - \sin\theta\,\hat{j}) d\theta \] Compute components: - \( \int_{\pi/2}^{3\pi/2} -\cos\theta\,d\theta = -[\sin\theta]_{\pi/2}^{3\pi/2} = -[\sin(3\pi/2) - \sin(\pi/2)] = -[-1 - 1] = 2 \) - \( \int_{\pi/2}^{3\pi/2} -\sin\theta\,d\theta = [\cos\theta]_{\pi/2}^{3\pi/2} = [\cos(3\pi/2) - \cos(\pi/2)] = [ - ] = \) So, \[ \vec{E}_{\text{arc}} = \frac{\rho_L}{4\pi\epsilon_ a} (2\hat{i}) = \frac{\rho_L}{2\pi\epsilon_ a} \hat{i} \] --- ## **Step 2: Semi-infinite Lines Contribution** For a **semi-infinite line** starting at \(x=\) and extending to \(x=+\infty\), parallel to x-axis, a perpendicular distance \(a\) from \(O\): - Each line produces a field at \(O\): - For line at \(y = +a\): \(dq = \rho_L dx\), position vector: \(x\hat{i} + a\hat{j}\) - Distance from \(O\): \(r = \sqrt{x^2 + a^2}\) - Direction from charge to \(O\): \(-x\hat{i} - a\hat{j}\) \[ d\vec{E}_1 = \frac{1}{4\pi\epsilon_} \frac{\rho_L dx}{(x^2 + a^2)} \frac{-(x\hat{i} + a\hat{j})}{\sqrt{x^2 + a^2}} \] \[ d\vec{E}_1 = -\frac{\rho_L}{4\pi\epsilon_} \frac{x\hat{i} + a\hat{j}}{(x^2 + a^2)^{3/2}} dx \] Integrate \(x\) from \(\) to \(+\infty\). \[ \vec{E}_1 = -\frac{\rho_L}{4\pi\epsilon_} \int_^{\infty} \frac{x\hat{i} + a\hat{j}}{(x^2 + a^2)^{3/2}} dx \] \(x\hat{i}\) term: \[ \int_^{\infty} \frac{x}{(x^2 + a^2)^{3/2}} dx = \quad \text{(odd function about } x=\text{, integrated from to }\infty\text{)} \] \(a\hat{j}\) term: \[ \int_^{\infty} \frac{a}{(x^2 + a^2)^{3/2}} dx \] Let \(x = a\tan\theta\), \(dx = a\sec^2\theta d\theta\), \(x^2 + a^2 = a^2\sec^2\theta\) When \(x=\): \(\theta=\), \(x=\infty\): \(\theta \to \pi/2\) \[ \int_^{\infty} \frac{a}{(x^2 + a^2)^{3/2}} dx = \int_^{\pi/2} \frac{a}{(a^2\sec^2\theta)^{3/2}} a\sec^2\theta d\theta \] \[ = \int_^{\pi/2} \frac{a^2\sec^2\theta}{a^3 \sec^3\theta} d\theta = \int_^{\pi/2} \frac{1}{a} \cos\theta d\theta \] \[ = \frac{1}{a} [\sin\theta]_^{\pi/2} = \frac{1}{a} (1 - ) = \frac{1}{a} \] So, \[ \vec{E}_1 = -\frac{\rho_L}{4\pi\epsilon_} \left( \cdot\hat{i} + \frac{1}{a}\hat{j} \right) = -\frac{\rho_L}{4\pi\epsilon_ a}\hat{j} \] Similarly, for the segment at \(y = -a\), the field at \(O\) is: \[ \vec{E}_2 = +\frac{\rho_L}{4\pi\epsilon_ a}\hat{j} \] So, the two straight segments **cancel out in the y-direction** and add nothing in the x-direction. --- ## **Step 3: Total Field at the Origin** \[ \boxed{ \vec{E}_{\text{total}} = \frac{\rho_L}{2\pi\epsilon_ a} \hat{i} } \] - **Direction:** Along the positive x-axis - **Magnitude:** \( \frac{\rho_L}{2\pi\epsilon_ a} \) --- ### **Summary Table** | Segment | Electric Field at O | Direction | |------------------|----------------------------|-------------------| | Semicircular arc | \( \frac{\rho_L}{2\pi\epsilon_ a} \) | \(+\hat{i}\) | | Upper line | \( -\frac{\rho_L}{4\pi\epsilon_ a} \) | \(-\hat{j}\) | | Lower line | \( +\frac{\rho_L}{4\pi\epsilon_ a} \) | \(+\hat{j}\) | | **Total** | \( \frac{\rho_L}{2\pi\epsilon_ a} \) | \(+\hat{i}\) | --- ## **Final Answer** \[ \boxed{ \vec{E}_O = \frac{\rho_L}{2\pi\epsilon_ a}\, \hat{i} } \] **The electric field at the origin is directed to the right (along the positive x-axis) with the magnitude shown above.**