This problem focuses on determining the electric field produced by a uniformly charged rod at specific points in space using calculus and the principles of electrostatics. Let's address each part.

Part (a): Derive an expression for the electric field $E_A$ at Point A

The task is to derive the magnitude of the electric field $E_A$ at Point A, which is along the $x$-axis at a distance $a$ from the end of the rod.

1. Charge distribution:
   The rod has a uniform linear charge density $\lambda_0 = \frac{q}{L}$, where $q$ is the total charge and $L$ is the rod's length.

2. Differential charge element $dq$:
   For a small segment of the rod of length $dx$, the charge is $dq = \lambda_0 dx$.

3. Electric field due to $dq$:
   The contribution of the field $dE$ from $dq$ at distance $r$ is:
   $dE = \frac{1}{4\pi \epsilon_0} \frac{dq}{r^2}$ Here, $r$ is the distance between $dq$ and Point A. Since $dq$ is located at position $x$ (along the rod), and Point A is at $-a$, the distance is $r = x + a$.

4. Integrate to find $E_A$:
   The total electric field at Point A is obtained by integrating along the length of the rod: $E_A = \int_0^L \frac{1}{4\pi \epsilon_0} \frac{\lambda_0 dx}{(x + a)^2}$ Factoring out constants: $E_A = \frac{\lambda_0}{4\pi \epsilon_0} \int_0^L \frac{dx}{(x + a)^2}$

Part (b): Write the integral for $E_B$ at Point B

Point B lies directly below the center of the rod along the $y$-axis at a distance $a$.

1. Geometry and field components:
   The symmetry of the setup simplifies the problem. The horizontal components of the electric field cancel out because the rod is centered, leaving only the vertical component of the electric field.

2. Vertical component of $dE$:
   For a small charge element $dq = \lambda_0 dx$ at a position $x$, the distance from $dq$ to Point B is:
   $r = \sqrt{x^2 + a^2}$ The vertical component of $dE$ is:
   $dE_y = \frac{1}{4\pi \epsilon_0} \frac{dq}{r^2} \cdot \frac{a}{r}$

3. Expression for $E_B$:
   Integrate over the entire length of the rod: $E_B = \int_{-\frac{L}{2}}^{\frac{L}{2}} \frac{1}{4\pi \epsilon_0} \frac{\lambda_0 a \, dx}{(x^2 + a^2)^{3/2}}$

Let me know if you'd like further clarification, step-by-step evaluation, or details on how to solve these integrals.

Relative Questions

1. What mathematical techniques can be used to evaluate the integral for $E_A$?
2. How does the distance $a$ affect the magnitude of $E_A$ and $E_B$?
3. Why do the horizontal components of the electric field cancel at Point B?
4. What is the physical significance of $\lambda_0$ in the problem?
5. How would the expressions change if the charge distribution were not uniform?

Tip

Always check for symmetry in electric field problems—it simplifies the calculation by eliminating unnecessary components.