2. 5/5 points | PreviouS Answers MI4 15.1.021 My Notes A thin rod lies on the x-
ID: 1772264 • Letter: 2
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2. 5/5 points | PreviouS Answers MI4 15.1.021 My Notes A thin rod lies on the x-axis with one end at -A and the other end at A, as shown in the diagram. A charge of -Q is spread uniformly over the surface of the rod. We want to set up an integral to find the electric field at location 0, Y, 0 due to the rod. Following the procedure discussed in the textbook, we have cut up the rod into small segments, each of which can be considered as a point charge. We have selected a typical piece, shown in red on the diagram (0,y,0) -0 -A Use the following as necessary: x, , dx, A, Q. Remember that the rod has charge -Q (a) In terms of the symbolic quantities given above and on the diagram, what is the charge per unit length of the rod? 24 (b) What is the amount of charge dQ on the small piece of length dx? ir (c) What is the vector from source to observation location? (d) What is the distance from the source to the observation location? (e) When we set up an integral to find the electric field at the observation location due to the entire rod, what will be the integration variable?Explanation / Answer
charge per unit length = total charge on the body / total length
total charge = -Q
total length = A + A
total length = 2A
charge per unit length = -Q / 2A
charge in segment x of rod = charge per unit length * length of the segment
if dx is the length of small segment then charge dQ in that segment will be
dQ = charge per unit length * dx
since charge per unit length = -Q / 2A so,
dQ = -Q * dx / 2A
horizontal distance of observation location from source = x units to negative x axis direction
vertical distance of observation location from source = y units in positive y axis direction
writing it in vector form
x component will be -xi as direction is toward negative x axis
y component will be yj as direction is toward positive y axis
vector from source to observation location r = -xi + yj
distance = sqrt(x component^2 + y component^2)
distance = sqrt((-x)^2 + y^2)
distance = sqrt(x^2 + y^2)
integration variable will be x since integration is the process of summation of the small segments here we broke length to small segments dx, to integrate it back to whole length we need equation in the form of x variable
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