# Uniformly charged rod electric field

A uniformly charged rod of length $L$ and total charge $Q$ lies along the $x$ axis as shown in in the figure. (Use the following as necessary: $Q$, $L$, $d$, and $k_e$.)

(a) Find the components of the electric field at the point $P$ on the $y$ axis a distance $d$ from the origin. $E_x=\,?$, $E_y=\,?$

I think I am getting close to a solution (maybe) but I keep getting turned around and confused. This is what I have so far. So, I know that I need $\lambda=Q/L$ and $r=\sqrt{d^2+L^2}$. My first thought was that the formula I need is $$E=\int_0^L \dfrac{\lambda \,k_e}{x^2}dx$$ but then it becomes a problem when I integrate and have to evaluate $\dfrac{-\lambda k_e}{x}$ at $x=0$. So then I tried to integrate the $x$ and $y$ components separately with respect to $\theta$. I found the following formulas at this site on the second slide.

$$E_x=\frac{-k_e\lambda}{L}(\sin\theta_2-\sin\theta_1)$$

$$E_y=\frac{-k_e\lambda}{L}(\cos\theta_2-\cos\theta_1)$$

Where, for this problem, $\theta_1=0$ and $\theta_2$ is the angle when x=L, so that $\tan^{-1}(\frac{d}{L})$, and this is where I really got lost. Are those formulas right for this problem? I had a thought to use the inverse sin and cos instead of the tan for $\theta_2$ but that made the formulas really messy. This is an online submission, so I'm guessing it should reduce to something nice, but I can't get to it.

• 1.) You certainly do not want to integrate $\int dx/x$. The denominator should be the distance between the source point and the field point. 2.) Electric field is a vector, and you integral has no vectors in it. 3.) Integrating the $x$ and $y$ is a good idea, but I don't know what you mean by "integrate ... with respect to $\theta$. – garyp Jun 9 '16 at 21:09
• @garyp So if I used $E=\int{_0^L \frac{ \lambda*k_e}{\sqrt{x^2+d^2}}dx}$ it should work? How do I add a vector into the integral? – Amanda Jun 9 '16 at 21:28
• by integrating with respect to $\theta$ I mean that I tried to integrate with $d\theta$ instead of $dx$ – Amanda Jun 9 '16 at 21:34
• Your integrand gives you the magnitude of the E-field for each $dx$. Actually, it doesn't. Take a close look at your integrand, and think about the quantity that you are calculating. (Purposefully vague hints.) – garyp Jun 9 '16 at 21:45
• @garyp I get that you're supposed to be vague and not just throw me the answer but you are confusing me so much more than if you would just walk me through it. I don't know what you're saying. Instead of doing the vector in the integral which may or may not be what you're talking about, can I split it in two integrals with $E_x=Esin(\theta)$ and $E_y=Ecos(\theta)$? And then $sin(\theta)=(L/R)$ and $cos(\theta)=(d/r)$ ? Am I not supposed to integrate from 0 to L? – Amanda Jun 9 '16 at 21:54

Often the easiest way to do such problems is using potential rather than the electric field. And always, it is best to use the appropriate coordinates. In this case, cylindrical coordinates with the rod at $\rho=0$ and extending from $z=-L/2$ to $z=+L/2$ are natural.

So what you do is say that for an element of rod with a tiny length $d\ell$ and charge $\frac{Q}{L}d\ell$ located at $0,z,\theta$ (where $\theta$ is actually moot since $\rho = 0$), the potential at any other point $(r,\zeta, \theta)$ os $$\frac{Q}{L\sqrt{r^2 + (z-\zeta)^2}}d\ell$$.

Then the potential due to the whole rod is $$V(r,\zeta, \theta) =\int_{z=-L/2}^{+L/2} \frac{Q}{L\sqrt{r^2 + (z-\zeta)^2}}d\ell$$

Then you do the integral, and when you get $V(r,\zeta, \theta)$ you take its gradient, because the negative of the gradient will be the electric field.

CAREFUL: The gradient operator in cylindrical coordinates is not quite just $(\frac{\partial}{\partial \rho},\frac{\partial}{\partial z},\frac{\partial}{\partial \theta})$. Learn what it is, then apply it to your $V$. And don't foregt to convert the field back to get $(E_x, E_y, E_z$ if your problem asks for those.