- #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 zero-point 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 zero-point 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!