The fleld dE due to this segment at the point P is in the negative x direction, and its magnitude is given by the following expression. Each element of the charge distribution produces a field at P in the negative x direction, so the vector sum of their contributions reduces to an algebraic sum. The total field at P due to all segments of the rod, which are at distances from P, Is given by Equation 23.11, which in this case becomes the following where the limits on the integral extend from one end of the rod (x-α) to the other (x-t+a). Because ke and λ are constants, they can be removed from the integral. Therefore, we find the following where we have used λ-Q/ 1 for the linear charge density. (Use the following as necessary: ke, Q, a, and t.) E-k,시스- To finalize, note that E decreases as a increases, as we expected from our mental representation. If point P is very far from the rod (a » t, we can ignore the in the denominator, and E- keQ/a2. This result is just the form you would expect for a point charge. Therefore, at large values of a, the charge distribution appears to be a point charge of magnitude Q as you should expect A rod 10.5 cm long is uniformly charged and has a total charge of-26.0 (a) Determine the magnitude of the electric field along the axis of the rod at a point 36.0 cm from its center. N/C (b) Determine the direction of the electric field along the axis of the rod at the same point. Away from the rod perpendicular to the axis of the rod Toward the rod along the axis of the rod D Toward the rod perpendicular to the axis of the rod Away from the rod along the axis of the rod