**Problem Statement:**Consider a rod of length \( L \) carrying a charge \( q \) distributed uniformly over its length.A diagram is shown with:- A horizontal rod of length \( L \).- A point \( P \) located at a distance \( a \) from the near end of the rod along the line extending from the rod.- The charge \( q \) is distributed along the rod.Where applicable, let \( V(r \to \infty) = 0 \).**Questions:****a.** What is the voltage \( V \) at point \( P \) (at distance \( a \) away from the near end of the rod) due to the charge over the length of the rod? Express your answer in terms of given parameters (\( L, q, a \)) and physical constants (\( k_e, \varepsilon_0 \)). Use underscore ("_") for subscripts and spell out Greek letters.*Hint for (a)*\[ V_P = \text{(Fill in the expression here)} \]**b.** Calculate the electric field at point \( P \) by differentiating \( V \) with respect to \( a \). Let positive sign of \( E \) indicate direction of electric field pointing away from the rod.*Hint for (b)*\[ E = \text{(Fill in the expression here)} \]**Interface:**- A button labeled "Hint."- A section to fill the answers for part (a) and part (b).- Options for "Question Help: Message instructor" and "Submit Question."

We are given the rod with uniform charge distribution. We need to find potential at a given point due to rod. The potential due to point charge is given as V=kqr Here q,r are charge and distance of point respectively. We first find small potential due to small element. We then integrate it. Charge is q. Length of rod is L. We take small length element dx. Charge density is λ=qL Charge on it is dq=λdx Potential due to this small charge element is dV=kdqx Total potential is V=∫dVV=∫L+aakdqxV=∫L+aakqdxLxV=kqL∫L+aadxxV=kqL[lnaL+a]The electric field is given as E=-∂V∂a Hence the field is E=-∂∂akqLlnaL+aE=-kqL∂∂alnaL+aE=-kqLL+aa∂∂aaL+aE=-kqLL+aaLL+a2E=-kqa(L+a)

Answered: Consider a rod of length I carrying a…