Certainly! Let’s break down the problem step by step, as you asked. --- ## **Step 1: Calculate the E-field at Point A (Bisector of Line of Charge)** **Given:** - Line of charge \(L = 2a\) units long - Uniform charge distribution - Point A is on the perpendicular bisector of the line (as shown in the diagram) ### **Approach:** 1. **Divide the line into infinitesimal segments** of length \(dx\), each carrying charge \(dq = \lambda dx\), where \(\lambda = \frac{Q}{L}\). 2. **Distance from any segment** at position \(x\) to point A (on bisector, at distance \(y\)) is \(r = \sqrt{x^2 + y^2}\). 3. **By symmetry,** only the vertical components of the E-field add up at the bisector; horizontal (x) components cancel. ### **Expression:** The electric field from a segment at point A: \[ dE = \frac{1}{4\pi\epsilon_} \frac{dq}{r^2} \] Vertical component: \[ dE_y = dE \cdot \frac{y}{r} \] Substitute \(dq = \lambda dx\) and \(r = \sqrt{x^2 + y^2}\): \[ dE_y = \frac{1}{4\pi\epsilon_} \frac{\lambda dx}{x^2 + y^2} \frac{y}{\sqrt{x^2 + y^2}} = \frac{1}{4\pi\epsilon_} \frac{\lambda y\, dx}{(x^2 + y^2)^{3/2}} \] **Integrate from \(-a\) to \(+a\):** \[ E_{y,A} = \int_{-a}^{a} \frac{1}{4\pi\epsilon_} \frac{\lambda y}{(x^2 + y^2)^{3/2}} dx \] **Explanation:** At the perpendicular bisector, by symmetry, the horizontal field components cancel, and only the vertical components add, leading to the above integral. --- ## **Step 2: Calculate the E-field at Point B (Along the Axis of Line of Charge)** **Given:** - Point B is along the axis of the line (let's call its distance from the line's center as \(y\)) ### **Approach:** 1. **Again, divide the line into infinitesimal segments** as before. 2. **Distance from segment at \(x\) to point B** is \(r = |y - x|\) (if B is along the axis, we need to set up coordinates appropriately). 3. **All field contributions are along the axis** (no cancellation). ### **Expression:** From a segment at \(x\): \[ dE = \frac{1}{4\pi\epsilon_} \frac{dq}{(y - x)^2} \] \[ dq = \lambda dx \] \[ E_{B} = \int_{-a}^{a} \frac{1}{4\pi\epsilon_} \frac{\lambda}{(y - x)^2} dx \] **Explanation:** Along the axis, all the field vectors point in the same direction, so you sum them directly. The result is an integral that's different from the bisector case. --- ## **Final Answer** - **At Point A (Bisector):** \[ E_{y,A} = \frac{\lambda y}{4\pi\epsilon_} \int_{-a}^{a} \frac{dx}{(x^2 + y^2)^{3/2}} \] - **At Point B (Axis):** \[ E_{B} = \frac{\lambda}{4\pi\epsilon_} \int_{-a}^{a} \frac{dx}{(y - x)^2} \] - Where \(\lambda = \frac{Q}{2a}\) and \(a\) is half the length of the line. --- **Summary:** - For each field point, divide the charge into tiny segments, sum their contributions using superposition, and write the integral for the electric field. - At the bisector, only vertical components add (symmetry). - On the axis, all contributions add directly. - Convert the sum to an integral as shown above. If you want explicit evaluation or numeric results, provide values for \(a\), \(y\), and \(Q\)!