- #1
Shinobii
- 34
- 0
Hello,
The electric potential is defined as:
$$
\phi(\vec{r}) = \int d^3r' \frac{\rho(\vec{r}')}{|\vec{r} - \vec{r}'|}.
$$
My question is, for solving for the potential inside of a charged solid sphere (constant charge density) by using the above equation I get,
$$
\frac{Q}{2R} \bigg( \frac{r^2}{R^2} - 3 \bigg).
$$
When the result is actually the above multiplied by a minus sign. Is this because when we take the reference point to be infinity, that we apply a minus sign by convention?
I just want to be sure exactly why the minus sign is introduced.
The electric potential is defined as:
$$
\phi(\vec{r}) = \int d^3r' \frac{\rho(\vec{r}')}{|\vec{r} - \vec{r}'|}.
$$
My question is, for solving for the potential inside of a charged solid sphere (constant charge density) by using the above equation I get,
$$
\frac{Q}{2R} \bigg( \frac{r^2}{R^2} - 3 \bigg).
$$
When the result is actually the above multiplied by a minus sign. Is this because when we take the reference point to be infinity, that we apply a minus sign by convention?
I just want to be sure exactly why the minus sign is introduced.