The Electric Field Due to a Charged Rod A rod of length e has a uniform positive charge per unit length A and a total charge Q. Calculate the electric field at a point P that is located along the long axis of the rod and a distance a from one end (see figure). The electric field at P due to a uniformly charged rod lying along the x-axis. P. SOLUTION Conceptualize The field de at P due to each segment of charge on the rod is in the Select vx-direction because every segment carries a positive charge. The figure shows the appropriate geometry. In our result, we expect the electric field to become smaller as the distance a becomes larger because point Pis farther from the charge distribution. Categorize Because the rod is continuous, we are evaluating the field due to a continuous charge distribution rather than a group of individual charges. Because every segment of the rod produces an electric field in the Select- V x-direction, the sum of their contributions can be handled without the need to add vectors. Analyze Let's assume the rod is lying along the x-axis, dx is the length of one small segment, and da is the charge on -Select- v. Because the rod has a charge per unit length A, the charge da on the small segment is da = 2 dx. Find the magnitude of the electric field at P due to one segment of the rod having a charge dg. (Use the following as necessary: a, k à, &, and x.) dx Find the total field at Pusing the equation for the electric field due to a continuous charge distribution: Noting that k, and A= are constants and can be removed from the integral, evaluate the integral. (Use the following as necessary: a, k , and Q.) (1) E =

The Electric Field Due to a Charged Rod A rod of length e has a uniform positive charge per unit length A and a total charge Q. Calculate the electric field at a point P that is located along the long axis of the rod and a distance a from one end (see figure). The electric field at P due to a uniformly charged rod lying along the x-axis. P. SOLUTION Conceptualize The field de at P due to each segment of charge on the rod is in the Select vx-direction because every segment carries a positive charge. The figure shows the appropriate geometry. In our result, we expect the electric field to become smaller as the distance a becomes larger because point Pis farther from the charge distribution. Categorize Because the rod is continuous, we are evaluating the field due to a continuous charge distribution rather than a group of individual charges. Because every segment of the rod produces an electric field in the Select- V x-direction, the sum of their contributions can be handled without the need to add vectors. Analyze Let's assume the rod is lying along the x-axis, dx is the length of one small segment, and da is the charge on -Select- v. Because the rod has a charge per unit length A, the charge da on the small segment is da = 2 dx. Find the magnitude of the electric field at P due to one segment of the rod having a charge dg. (Use the following as necessary: a, k à, &, and x.) dx Find the total field at Pusing the equation for the electric field due to a continuous charge distribution: Noting that k, and A= are constants and can be removed from the integral, evaluate the integral. (Use the following as necessary: a, k , and Q.) (1) E =