Example 23.7 The Electric Field Due to a Charged Rod Problem A rod of length has

Question

Example 23.7 The Electric Field Due to a Charged Rod Problem A rod of length has a uniform linear charge density ? and a total charge Q. Calculate the electric field at a point P along the axis of the rod, a distance a from one end (Fig. 23.17) Strategy Choose a small element of the charge distribution, such as the blue section marked in the diagram. Find the electric field due to the segment, then sum up all of the segments' vector contributions to the overall field at a point P along the axis of the roc. Figure 23.17 The electric field at P due to a uniformly charged rod lying along the x axis. The field at P due to the segment of charge dq is k2dq/x2. The total field at P is the vector sum over all segments of the rod. Solution Figure 23.17 helps us visualize the source of the electric field and conceptualize what the field might look like We expect the field to be symmetric around the horizontal dimension of the rod and would expect the field to decrease for increasing values of a. We categorize this problem as one involving a continuous distribution of charge on the rod rather than a collection of individual charges. To analyze the problem, we choose an infinitesimal element of the charge distribution as indicated by the blue portion in Figure 23.17. Let us use dx to represent the length of one small segment of the rod and let dq be the charge on the segment. We express the charge dq of the element in terms of the other variables within the integral (in this example, there is one variable, x). The charge dq on the small segment is dq = ? dx. The field dE due to this segment at the point P is in the negative x direction, and its magnitude is given by the following expression. dq, Xdx 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 different 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 - a) to the other (x - ! + a)

Explanation / Answer

Execise 23.7)

Ans)

Given data is charge q=-22.5?C

D=0.25-0.11=0.14m

L=0.11m

D+L=0.14+0.11=0.25

E=kq/[D(D+L)]

E=(9x109x-22.5x10-6)/(0.14x0.25)

E=-5.78x106 C/m2

E is ponting towards the center of the rod

Exercise 23.8)

Ans)The electric field along the axis of a uniformly charged ring is given by

Given q=71.5?c,r=14cm

a)

E=kqx/(x2+r2)3/2

=(9x109x71.5x10-6x3)/(9+196)3/2

=79.27 N/C

b)x=17cm

E=(9x109x71.5x10-6x17)/(289+196)3/2

=1024.2N/C

c)x=100cm

E=( 9x109x71.5x10-6x100)/(10000+196)3/2

=62.5N/C

Related Questions

Navigate

• Browse (All)
• Browse E
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.