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Introductory Physics Homework Help
Voltage using different references
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[QUOTE="darioslc, post: 6389873, member: 681636"] [B]Homework Statement:[/B] How the choice of zero voltage at the origin changes the electric field. [B]Relevant Equations:[/B] ##\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! [/QUOTE]
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Introductory Physics Homework Help
Voltage using different references
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