In the figure below, a plastic rod of length 2L is shown with a charge of q. Poi
ID: 1655743 • Letter: I
Question
In the figure below, a plastic rod of length 2L is shown with a charge of q. Point P is centered along the line of the midpoint of this rod. SHOW ALL WORK FOR FULL CREDIT. a. Draw a Free Body Diagram on the figure to the left showing the representative electric field acting at point P from a small incremental segment at some point off the midpoint axis of the rod. Assign variables for the x-axis distance to point p and the y-axis distance to your small increment location b. Write an expression for the linear charge density and how it can be altered to represent the charge fraction on a small increment of the rod. Use the coordinate system shown in the diagram. c. Show how the electric field expression of for a point charge can be modified to represent these incrementally small sections of charged rod by writing the expression first with dq and charge density expression of part (b) above. d. Show how this field expression from part (c) is separated into x-and y-axis components based on your free body diagram. Rewrite this expression in terms of y (or x...) rather than r and theta. Explain any symmetry considerations that may simplify this problem. e. Integrate to answer the question: What is the total electric field at point P due to the full charge on the rod? Show your answer in simplified form in symbolic notationExplanation / Answer
a. FBD
b. linear charge density can be written as lambda = Q/2L [ where Q is charge on the rod]
charge fraction on small part in figure, dq = lambda*dy
c. electric field at P = k*dq/(d^2 + y^2)
dE = k*lambda*dy/(d^2 + y^2)
d. dE is a vector when seperated into components can be written as
dEx = dEcos(theta) = dE*d/sqroot(d^2 + y^2) = k*lambda*d*dy/(d^2 + y^2)^3/2
dEy = dEsin(theta) = dE*y/sqroot(d^2 + y^2) = k*lambda*y*dy/(d^2 + y^2)^3/2
e. consider
dEx = dEcos(theta) = dE*d/sqroot(d^2 + y^2) = k*lambda*d*dy/(d^2 + y^2)^3/2
integrating we get
Ex = k*lambda[y/sqroot(y^2 + d^2)]/d [ from y = -l to +l
applygin limits
Ex = 2k*lambda[l/sqroot(l^2 + d^2) ]/d
similiarly
dEy = dEsin(theta) = dE*y/sqroot(d^2 + y^2) = k*lambda*y*dy/(d^2 + y^2)^3/2
now this is an odd funciton of y and integrating it from -l to +l will give 0
so Ey = 0