Voltage using different references

[darioslc]

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!

 

[Master1022]

I believe potential is always measured relative to something, whereas the electric field is 'absolute'/fixed. Therefore, when you set the zero potential point to a different place, you can get a different potential as you are measuring relative to a different point. An electric circuit is an example; when we say the voltage is '5 V', we really mean that it has a potential that is +5V greater than the chosen zero (i.e. ground). If we chose the ground elsewhere, then the potential would be different I think.

 

[Steve4Physics]

In both cases, when I apply the gradient $\vec{E}=-\nabla V$ I get the same field

Hi. You are supposed to get the same field. Changing the zero potential reference point does not affect the field.

Changing the reference point simply adds a constant amount to every point's potential.

A change from 100V to 150V over a some distance gives the same field as a change from 120V to 170V over the same distance.

Think of the equivalent gravitational situation: g = -9.81m/s² at the Earth's surface whether you choose to take V=0 at ground level or V=0 at ∞.

 

[darioslc]

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

 

[Steve4Physics]

That's right - the field cannot be changed by changing the reference point. Note that this in not limited to spherical charge distributions, it is always true. I guess the question was set to make you think about why it is true.

If we change the zero reference point, the potential at every point changes by the same fixed amount (say k). Since constants disappear on differentiation:
$$\vec{E}=-\nabla (V+k) = -\nabla (V)$$