Finding electric field 3 cm along a thin rod.

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SUMMARY

The discussion focuses on calculating the electric field at a point 3 cm along a thin rod of length 5.0 cm and total charge 8.4 nC. The correct formula for the electric field due to a line of charge is derived as E = KQ/(d*(d^2 + (L/2)^2)^(1/2)), where K is the Coulomb's constant (8.99 x 10^9 N m²/C²). The user initially misapplied the formula, resulting in an incorrect value of 64458.88 N/C, while the correct calculation yields an electric field of approximately 2.7 x 10^5 N/C using the appropriate integration method.

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Homework Statement


The figure shows a thin rod of length L = 5.0 cm with total charge Q = 8.4 nC. What is the magnitude of the electric field E at x = 3.0 cm?

Figure in the attachment

Homework Equations


E = Kq/r^2
K = 8.99 * 10^9 N m^2/C^2;
r is the distance between the point and the charge.
Field for a line of charge:
E = KQ/(d*(d^2 + (L/2)^2)^(1/2))

The Attempt at a Solution


We have Q, L and "d". But I am still unable to get the answer.
When I plug in the values as they are (as what I think they are), I get 64458.88
Answer is: 2.7*10^5 N/C.

But, when I use linear charge density lambda = Q/L.
E = lambda/4*pi*epsilon_0 * distance, I get 2.2634 N/C. Do you think this is the right way to get it?
I am not sure why that equation does not work.
 

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Field due to a small charge element at the point is given by
dE = k*dQ/r^2.
dQ = Q/L*dr
Hence E = Integration(dE) = k*Q/L*Int(dr/r^2).
Find the lower and upper limit of r,Substitute it in the integration to get the field.
 

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