Electric field due to a uniformly charged rod

[sprinks13]

hi, here's the question:

a rod in the xy plane has it's ends at (0,0) and (L,0). It has a uniform charge per unit length (lambda). Find the electric field on the x-axis for 0 < x < L.

solution attempt:

dEx = dE
= kdQ/r^2

dQ = lambda*ds = (Q/L)*dx

now i took my r = (L - a - x) where x is the location of dx and a is the distance from L to the point that the field is being measured (note that 0 < a < L).

dEx = KQ/L (integral) dx/(L - a -x)^2

So my problem is my r. It's wrong, but I am not sure why or how to fix it.

thanks in advance!

 

[Doc Al]

now i took my r = (L - a - x) where x is the location of dx and a is the distance from L to the point that the field is being measured (note that 0 < a < L).

If a is the location where you want the field, then r is just the distance between a and x. Hint: break the region 0 to L into two: 0 to a & a to L.

 

[kerimek]

I would suggest looking into the formula for electric field due to a uniformly charged rod, which is given by:

E = (λ/2πε0)(1/r1 - 1/r2)

where λ is the linear charge density, ε0 is the permittivity of free space, r1 is the distance from the point on the x-axis to the end of the rod at (0,0), and r2 is the distance from the point on the x-axis to the end of the rod at (L,0).

Using this formula, you can calculate the electric field at any point on the x-axis between (0,0) and (L,0). It is important to note that the distance from the point on the x-axis to the rod is not simply r = (L-a-x), as the rod is not a point charge but a continuous distribution of charge. Instead, you will need to consider the distance from the point on the x-axis to each end of the rod separately.

I would also suggest checking your units and making sure that all values used in the formula are in SI units. Additionally, double check your integration limits to ensure that you are integrating over the entire length of the rod.

I hope this helps and good luck with your solution!