Let the rod be on the x-axes with endpoints -L/2 and L/2 and uniform charge density lambda (2.6nC/0.4m = 7.25 nC/m).
The point then lies on the y-axes at d = 0.03 m.
from symmetry, the field at that point will be ascending along the y-axes.
A charge element at position x on the rod has distance sqrt(x^2 + d^2) to the point.
Also, from the geometry, the component in the y-direction is d/sqrt(x^2+d^2) times the field strength.
All in all, the infinitesimal field strength from the charge between x and x+dx is:
dE = k lambda dx * 1/(x^2+d^2) * d/sqrt(x^2+d^2)
Therefore, upon integration,
E = k lambda d INTEGRAL{dx / (x^2 + d^2)^(3/2) } where x goes from -L/2 to L/2.
This gives:
E = k lambda L / (d sqrt((L/2)^2 + d^2) )
But lambda L = Q, the total charge on the rod, so
E = k Q / ( d * sqrt((L/2)^2 + d^2) )
<h2>
Answer: Toward the center of the circle.</h2>
This situation is characteristic of the uniform circular motion , in which the movement of a body describes a circumference of a given radius with constant speed.
However, in this movement the velocity has a constant magnitude, but its direction varies continuously.
Let's say 
is the velocity vector, whose direction is perpendicular to the radius 
of the trajectory, therefore
the acceleration 
is directed toward the center of the circumference.
32? I could be wrong but I’m going with that answer choice
