 #1
darioslc
 2
 0
 Homework Statement:
 How the choice of zero voltage at the origin changes the electric field.
 Relevant Equations:

##\oint_S\vec{E}\cdot d\vec{S}=\frac{Q}{\varepsilon_0}##
##V(r)=\int \vec{E}\cdot d\vec{l}##
The problem is for a solid sphere uniformly charged with Q and radii R.
First I calculated taked ##V(\infty)=0##, giving me for :
$$
\begin{align*}
V(r)=&\frac{3Q}{8\pi\varepsilon_0 R}\frac{Q}{8\pi\varepsilon_0 R^3}r^2\qquad\text{if $r<R$}\\
V(r)=&\frac{Q}{4\pi\varepsilon_0 r}\quad\text{if $r\geq R$}\\
\end{align*}
$$
so far well, but when I calculated the voltage with ##V(0)=0## I get a little similar expression:
$$
\begin{align*}
V(r)=&
\begin{cases}
\frac{Q}{8\pi\varepsilon_0 R^3}r^2\qquad\text{if $r<R$}\\
\frac{3Q}{8\pi\varepsilon_0 R}+\frac{Q}{4\pi\varepsilon_0 r}\qquad\text{if $r\geq R$}\\
\end{cases}
\end{align*}
$$
for both, I used the expression of the electric field
$$
\begin{align*}
\vec{E}(r)=&
\begin{cases}
\frac{Q}{4\pi\varepsilon_0R^3}r^2\qquad\text{if $r<R$}\\
\frac{Q}{4\pi\varepsilon_0r}\qquad\text{if $r\geq R$}\\
\end{cases}
\end{align*}
$$
In both cases, when I apply the gradient ##\vec{E}=\nabla V## I get the same field, and I can't understand how can change the field if I take other zeropoint references, is not independent of potential? ie, always I get the same field
Maybe have an error in calculus, but I didn't found it.
Thanks a lot!
First I calculated taked ##V(\infty)=0##, giving me for :
$$
\begin{align*}
V(r)=&\frac{3Q}{8\pi\varepsilon_0 R}\frac{Q}{8\pi\varepsilon_0 R^3}r^2\qquad\text{if $r<R$}\\
V(r)=&\frac{Q}{4\pi\varepsilon_0 r}\quad\text{if $r\geq R$}\\
\end{align*}
$$
so far well, but when I calculated the voltage with ##V(0)=0## I get a little similar expression:
$$
\begin{align*}
V(r)=&
\begin{cases}
\frac{Q}{8\pi\varepsilon_0 R^3}r^2\qquad\text{if $r<R$}\\
\frac{3Q}{8\pi\varepsilon_0 R}+\frac{Q}{4\pi\varepsilon_0 r}\qquad\text{if $r\geq R$}\\
\end{cases}
\end{align*}
$$
for both, I used the expression of the electric field
$$
\begin{align*}
\vec{E}(r)=&
\begin{cases}
\frac{Q}{4\pi\varepsilon_0R^3}r^2\qquad\text{if $r<R$}\\
\frac{Q}{4\pi\varepsilon_0r}\qquad\text{if $r\geq R$}\\
\end{cases}
\end{align*}
$$
In both cases, when I apply the gradient ##\vec{E}=\nabla V## I get the same field, and I can't understand how can change the field if I take other zeropoint references, is not independent of potential? ie, always I get the same field
Maybe have an error in calculus, but I didn't found it.
Thanks a lot!