Ch23: A charge of 3.00 11C is located at the top location a distance d = 11 mm, a 2.50 uC located at the origin and 2.00 uC located to the right along X-axis a distance d = 12.2 mm. Calculate a) the mid-horizontal line of the arrangement, then b) imagine a reference point P along that line a distance of 16.00 mm and calculate the force at that point, c) calculate the E-eld at that point P d) the AV at the same point. (50 pts) :! r! A rod of length & has a uniform positive charge per unit length > 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 (Fig. 23.2). Figure 23.2 The electric field at P due to a uniformly charged rod lying along the a axis. We choose the location of point P to be the origin. dae P Solution Conceptualize The field dE at P due to each segment of charge on the rod is in the negative a direction because every segment carries a positive charge. Figure 23.2 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 negative a direction, the vector sum of their contributions is easy to determine.Analyze Let's assume the rod is lying along the :1: axis, (is: is the length of one small segment, and dig is the charge on that segment. Because the rod has a charge per unit length A, the charge dq on the small segment is dq = A 03w. Find the magnitude of the electric eld at P due to (if? = k E = k A d3 one segment of the rod having a charge dq: e m2 e $2 Find the total eld at P using 7*. Equation 23.1: E : jam k8 Ad_:: :1? Noting that ice and A = QM are constants and can be removed from the integral, evaluate the integral