Recent content by darioslc

Yes, I'm sorry is early morning here and a bit tired. Finally I think I already understood. Thanks a lot

Hello, may be there are mistakes I will check the calculus (but I do not find errors). Sorry by confusion and delay in response, ##R## is the ##L## that I had put in the first drawing. The limits of integration was: $$\begin{align*}...

Yes diverge but, is it because the point y=0 is in the same line that wires? ie, the electric field close to (0,0) is very extreme (ideally), I add the graph that illustrate both

Thanks! finally I do using definition of potential: $$V(x,y)=\frac1{4\pi\varepsilon_0}\int\frac{dq}{|\vec{r}-\vec{r'}|}$$ in ##(0,y)## $$V(0,y)=\frac1{4\pi\varepsilon_0}\lambda\int\frac{1}{\sqrt{x'^{2}+y^{2}}}dx'$$ adding two potential due to wires...

Hello! I do it for definition because Gauss's law not possible. I use cartesian coordinates: $$dq=\lambda dx'\text{because is only in x direction}$$ $$\vec{r'}=(x',0)$$ $$\vec{r}-\vec{r'}=(x-x',y)$$ $$\hat{r}=\frac{(x-x',y)}{\sqrt{(x-x')^{2}+y^{2}}}$$...

But when you put the charge -Q in the opposite corner (when in original scheme it's absent charge), the field is the same but not the sense, no? The scheme with the center (0,0) is: 2q *----------------- | | | (0,0) | | | -Q...

Hello, I reasoned by simmetry, the two charges with value 2q not contributed at field because there are equidistant at point and are similar charges. Therefor only survival the field due to Q, using the definition for electric field of the puntual charge...

Hi, thanks for your responses. Then the field shouldn't vary? this is what I was thinking, or the question is a little captious.

The problem is for a solid sphere uniformly charged with Q and radii R. First I calculated taked ##V(\infty)=0##, giving me for : $$ \begin{align*} V(r)=&\frac{3Q}{8\pi\varepsilon_0 R}-\frac{Q}{8\pi\varepsilon_0 R^3}r^2\qquad\text{if $r<R$}\\ V(r)=&\frac{Q}{4\pi\varepsilon_0 r}\quad\text{if...