Calculate the electric field due to a line of charge of finite length

[Yalanhar]

Homework Statement

A thin rod of length L and charge Q is uniformly charged, so it has a linear charge
density $\lambda =q/l$ Find the electric field at point where is an arbitrarily positioned
point.

Relevant Equations

$dE=\frac{Kdq}{r^2}$

Homework Statement: A thin rod of length L and charge Q is uniformly charged, so it has a linear charge
density $\lambda =q/l$ Find the electric field at point where is an arbitrarily positioned
point.
Homework Equations: $dE=\frac{Kdq}{r^2}$

A thin rod of length L and charge Q is uniformly charged, so it has a linear charge
density $\lambda =q/l$ Find the electric field at point P where P is an arbitrarily positioned
point.

Tipler solves this one. But there are 2 things that I do not understand about his solution.

1) Why on 5 he used the negative sign for Xs? shouldn't he use Xs positive as in:
$tan\theta =\frac {y}{x}$
$x = \frac{y}{tan\theta}$
$dx = -ycsc^2\theta$

But by doing this my Ex gets negative.
2) What the negative sign on Ey means? The orientation is opposite to the unit vector j?

Sorry for bad image and english.

 

[TSny]

Yes, dealing with the signs in this calculation can be annoying.

1. Note that $\theta$ is between 0 and $\pi/2$. So, $\tan \theta$ is positive. But $\frac {y_p}{x_s}$ is negative (why?). So, it would not be correct to write $\tan \theta = \frac{y_p}{x_s}$.

2. In the final expression for $E_y$, there is an overall negative sign. However, you also need to consider the sign of the expression inside the parentheses.

 

[Yalanhar]

(1) Yes, but x is positive or negative? Tipler used negative so the derivative can be positive. But shouldn't i use modulus?

(2) I see it now. $cos\theta_{2} - cos\theta_{1}$ is negative. So double negative gives me positive.

 

[TSny]

(1) Yes, but x is positive or negative? Tipler used negative so the derivative can be positive. But shouldn't i use modulus?

I'm not sure I follow you here.

Do you agree with Tipler's expression $\tan \theta = \frac{y_p}{|x_s|}$?

If so, then note that Tipler's choice of the coordinate system implies that $x_s$ is a negative number. So, $|x_s| = -x_s$.

 

[Yalanhar]

I'm not sure I follow you here.

Do you agree with Tipler's expression $\tan \theta = \frac{y_p}{|x_s|}$?

If so, then note that Tipler's choice of the coordinate system implies that $x_s$ is a negative number. So, $|x_s| = -x_s$.

For example,

in this case $\theta = arctan(\frac {0.5}{0.87})$ and not $\theta = arctan(\frac {0.5}{(-0.87)})$

$\tan \theta = \frac{y_p}{|x_s|}$?I agree

$|x_s| = -x_s$. is this true because it's implied that x is negative, then he used -x to be positive? It makes sense

 

[TSny]

Yes. Whenever you have a quantity $x$ that is negative, then you can write $|x| = -x$. Both sides of this equation are positive numbers.

 

[Yalanhar]

Yes. Whenever you have a quantity $x$ that is negative, then you can write $|x| = -x$. Both sides of this equation are positive numbers.

Thanks ! i finally get it