lalllaay, Fundamentals of Physics, 10e In Figure (a), a circular plastic rod wit

Question

lalllaay, Fundamentals of Physics, 10e In Figure (a), a circular plastic rod with uniform charge +Q rods, each with identical uniform charges+Q, are added until the dirde is complete. A fifth arrangement (which would be labeled e of curvature (at the origin). In Figure (b), Figure (c), and Figure (d), labeled e) is like that ind fourth quadrant has charge -Q. except the rod in the id Rank the five arrangements according to the magnitude of the electric field at the center of curvature, greatest first. If multiple arrangements rank equally, use the same rank for each, then exclude the intermediate ranking (i.e. if objects A, B, and C must be ranked, and A and B must both be ranked first, the ranking would be A:1, B:1, C:3). If all arrangements rank equally, rank each as '1

Explanation / Answer

Electric field at origin due to the plastic charged rod of charge +Q in figure 1 be E1
so, consider a point on the rod, at angle theta with the x axis, with angular widtth d(theta)
then d(theta ) *R = dl [ dl isl ength of the segment of arc, and R is radius of the circle]
now, charge on this small part = dq = 2Q*dl/pi*R = 2Q*d(theta)/pi
electric field at origin due to this charge = kdq(-cos(theta)i -sin(theta)j)/R^2 = 2kQ(-cos(theta)i -sin(theta)j)d(theta)/pi*R^2

for case 1, dE has to be integrated form theta = 0 to theta = 90
E1 = integrate[2kQ(-cos(theta)i -sin(theta)j)d(theta)/pi*R^2] = 2kQ[(sin(90)i -cos(90)j) - (sin(0)i -cos(0)j)]/pi*R^2 = 2kQ[(i + j)]/pi*R^2
|E1| = 2kQ*sqroot(2)/pi*R^2

for case 2, dE has to be integrated form theta = 0 to theta = 180
E2 = integrate[2kQ(-cos(theta)i -sin(theta)j)d(theta)/pi*R^2] = 2kQ[(sin(180)i -cos(180)j) - (sin(0)i -cos(0)j)]/pi*R^2 = 4kQ[(j)]/pi*R^2
|E2| = 4kQ/pi*R^2

for case 3, dE has to be integrated form theta = 0 to theta = 270
E3 = integrate[2kQ(-cos(theta)i -sin(theta)j)d(theta)/pi*R^2] = 2kQ[(sin(270)i -cos(270)j) - (sin(0)i -cos(0)j)]/pi*R^2 = 2kQ[(-i + j)]/pi*R^2
|E3| = 2kQ*sqroot(2)/pi*R^2

for case 4, dE has to be integrated form theta = 0 to theta = 360
E4 = integrate[2kQ(-cos(theta)i -sin(theta)j)d(theta)/pi*R^2] = 2kQ[(sin(360)i -cos(360)j) - (sin(0)i -cos(0)j)]/pi*R^2 = 0
|E4| = 0

for case 5, dE has to be integrated form theta = 0 to theta = 360 but in parts
E5 = integrate[2kQ(-cos(theta)i -sin(theta)j)d(theta)/pi*R^2] = 2kQ[(sin(270)i -cos(270)j) - (sin(0)i -cos(0)j)]/pi*R^2 - 2kQ[(sin(360)i -cos(360)j) - (sin(270)i -cos(270)j)]/pi*R^2
E5 = 2kQ[(-i +j)]/pi*R^2 - 2kQ[( -j +i )]/pi*R^2 = 4kQ[(-i +j)]/pi*R^2
|E5| = 4kQ*sqroot(2)/pi*R^2

so |E1| : |E2| : |E3| : |E4| : |E5| = 2.828 : 4 : 2.828 : 0 : 5.65

so, E5 > E2 > E1 = E3 > E4
so E1 -> third greates
E2 -> second greatest
E3 -> Fourth greatest
E4 -> Least
E5 -> Greatest

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