Linear Charge Distribution Question

[im.zaur]

Homework Statement

Find the electric field a distance z above the midpoint of a straight line segment of length 2L, which carries a uniform line charge $$\lambda$$.

Homework Equations

dE = 2 * (1 / 4$$\pi\epsilon$$) * ($$\lambda$$dx / r²) * cos$$\theta$$ $$\hat{z}$$

cos$$\theta$$ = z/r

r = [tex]\sqrt{x² + z²}[/tex]

The Attempt at a Solution

This an example question from my textbook with a full solution. The problem I am having is that I don't understand where the x² disappears right after this step.

E = 1 / (4$$\pi$$$$\epsilon$$ $$\int$$ (2$$\lambda$$z) / ([tex]\sqrt[3]{x² + z²}[/tex]) dx

And I am not sure why the superscripts didn't work inside the square root function. Any help would be appreciated if anyone understands what I am asking for.

Thanks

 

[mighty2000]

for posting this question! It's great to see students actively seeking to understand their textbook problems.

First, let's clarify the notation in the equation you provided. The "dE" in the first equation should actually be "dEz", as you are finding the electric field in the z-direction. Additionally, the "r2" in the equation should actually be "r^2", indicating that it is squared.

Now, to address your confusion about the "x^2" term disappearing: in the integral, you have the term "dx" at the end. This indicates that you are integrating with respect to x. In this case, the "x^2" term is simply a constant factor and can be pulled out of the integral. This is a common technique in integration known as "pulling out a constant".

So, the integral can be rewritten as:

E = (2λz) / (4πε) * ∫ (1 / √(x^2 + z^2)) * dx

= (2λz) / (4πε) * x * ∫ (1 / √(x^2 + z^2)) * dx

= (2λz) / (4πε) * x * ln(√(x^2 + z^2)) + C

= (λz) / (2πε) * x * ln(x^2 + z^2) + C

Finally, we can evaluate the integral at the limits of integration (x = -L and x = L) to get the final solution:

E = (λzL) / (πε) * ln(L^2 + z^2)

I hope this helps clarify the solution for you. As for the superscripts not working inside the square root function, it could be due to a formatting issue or a mistake in the original equation. I would recommend double-checking the equation in your textbook to make sure all the superscripts are in the correct place.