Final answer:
To find the x component of the electric field at y = 2 m due to a uniform linear charge distribution along the x axis, one must integrate the contributions of the x-components of the electric field from each segment of the charged line, considering the symmetry and the finite size of the charge distribution.
Step-by-step explanation:
The student's question pertains to the calculation of the electric field due to a uniform linear charge distribution. Since the charge distribution is along the x-axis and we are looking for the electric field at a point on the y-axis, we can expect symmetry to play a role in the calculation.
For a uniform linear charge distribution λ (lambda), the electric field at a point away from the distribution can be found by integrating the contributions of each infinitesimal part of the charged object, using Coulomb's law. The component of the electric field in the x direction (Ex) due to a small segment of charge is given by Ex = (k dq cos(θ)) / r², where k is Coulomb's constant, dq is the charge of the segment, r is the distance to the point where the electric field is measured, and θ is the angle between the line joining the point and the segment, and the y-axis.
Due to symmetry, we can say that the x-components of the electric field from symmetric parts of the linear charge distribution will cancel each other out. Hence if this distribution were infinitely long, the net electric field along the x-direction at any point on the y-axis would indeed be zero. However, since the charge distribution is from x = 0 to x = 3 m, we need to take into account the finite size of the charge distribution. To find the x-component of the electric field at y = 2 m, we would integrate the x-component of the electric field due to each segment from x = 0 to x = 3 m, taking into account the varying angle θ and the distance r for different segments of charge.