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.dxPaSOLUTIONConceptualize The field dE at P due to each segment of charge on the rod is in the --Select-- vbecause point P is farther from the charge distribution.v x-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 largerCategorize 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--- vy x-direction, the sum of their contributions can be handledwithout 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 dg is the charge on --Select--- vBecause the rod has a charge per unit length a, the charge dg on the small segment is dg = a 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, 1, e, and x.)dxdadE = k=x²Find the total field at P using the equation for the electric field due to a continuous charge distribution:--?-- vE =Noting that k. and 1 = 2 are constants and can be removed from the integral, evaluate the integral. (Use the following as necessary: a, k, e, and Q.)-7- v-2-- vE - kA1(1)E =Finalize We see that our prediction is correct; if a becomes larger, the denominator of the fraction grows larger, and E--Select-. On the other hand, if a → 0, which corresponds to sliding the bar to the left until its left end is at the origin, then E - o. That represents the condition inwhich the observation point Pis at zero distance from the charge at the end of the rod, so the field becomes infinite.EXERCISEA rod 17.0 cm long is uniformly charged and has a total charge of –25.0 µC.Hint(a) Determine the magnitude of the electric field (in N/C) along the long axis of the rod at a point 37.0 cm from its center.

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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 ais. P SOLUTION Conceptualize The field dE at P due to each segment of charge on the rod is in the --Select- v x-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 P is 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 2, the charge da on the small segment is da - i 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, 2, e, and x.) ) dx Find the total field at Pusing the equation for the electric field due to a continuous charge distribution: E- Noting that k and - are constants and can be removed from the integral, evaluate the integral. (Use the following as necessary: a, k, , and Q.) (1) E- k- Finalize We see that our prediction is correct; if a becomes larger, the denominator of the fraction grows larger, and E--Select- which the observation point Pis at zero distance from the charge at the end of the rod, so the field becomes infinite. On the other hand, if a-0, which corresponds to sliding the bar to the left until its left end is at the origin, then E-. That represents the condition in EXERCISE A rod 17.0 cm long is uniformly charged and has a total charge of -25.0 pc. Hint (a) Determine the magnitude of the electric field (in N/C) along the long axis of the rod at a point 37.0 cm from its center.