Question Solved1 Answer The figure (figure 1) shows a thin rod of length L with total charge Q. Find an expression for the electric field strength on the axis of the rod at distance r from the center. Express your answer in terms of the variables L, Q, r, and appropriate constants. Verify that your expression has the expected behavior if r >> L. Express your answer In terms of variables Q, r and constants pi, epsilon. Evaluate E at r = 3.4 cm if L = 5.0 cm and Q = 3.8 nC. Express your answer to two significant figures and Include the appropriate units.

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Transcribed Image Text: The figure (figure 1) shows a thin rod of length L with total charge Q. Find an expression for the electric field strength on the axis of the rod at distance r from the center. Express your answer in terms of the variables L, Q, r, and appropriate constants. Verify that your expression has the expected behavior if r >> L. Express your answer In terms of variables Q, r and constants pi, epsilon. Evaluate E at r = 3.4 cm if L = 5.0 cm and Q = 3.8 nC. Express your answer to two significant figures and Include the appropriate units.
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Transcribed Image Text: The figure (figure 1) shows a thin rod of length L with total charge Q. Find an expression for the electric field strength on the axis of the rod at distance r from the center. Express your answer in terms of the variables L, Q, r, and appropriate constants. Verify that your expression has the expected behavior if r >> L. Express your answer In terms of variables Q, r and constants pi, epsilon. Evaluate E at r = 3.4 cm if L = 5.0 cm and Q = 3.8 nC. Express your answer to two significant figures and Include the appropriate units.
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Suppose you need to calculate the electric field atpoint P located along the axis of a uniformly charged rod. Let thecharge distribution per unit length along the rod be represented byl; that is, lambda=(Delta q)/(Delta x). The net chargerepresented by the entire length of the rod could then be expressedas Q = lL.Let's first combine F = qE and Coulomb's Law F=(kqQ)/(r^(2))=(1)/(4piepsi_(0))(qQ)/(r^(2)) to derive anexpression for E.E=(F)/(q)E=((1)/(4piepsi_(0))(qQ)/(r^(2)))/(q)E=(1)/(4piepsi_(0))(Q)/(r^(2))For a small section Dxi along the rod, thecharge present will be Dqi whereDqi = lDxiThe electric field contribution at P by this sectionwould be represented byDeltaE_(i)=(1)/(4piepsi_(0))(Deltaq_(i))/(x_(i)^(2))The total magnitude of the electric field at P would beequal to the sum of all these smaller contributions,DEi.E=sum DeltaE_(i)E=sum(1)/(4piepsi_(0))(Deltaq_(i))/(x_(i)^(2))E=(1)/ ... See the full answer