Answer:
- No, this doesn't mean the electric potential equals zero.
Step-by-step explanation:
In electrostatics, the electric field
is related to the gradient of the electric potential V with :
![\vec{E} (\vec{r}) = - \vec{\\abla} V (\vec{r})](https://img.qammunity.org/2020/formulas/physics/college/k3an340oips7ylsvwfth7r9utv39l1poe1.png)
This means that for constant electric potential the electric field must be zero:
![V(\vec{r}) = k](https://img.qammunity.org/2020/formulas/physics/college/ah8k2zqisa6lhuv53mq5pdxugonyp70rvh.png)
![\vec{E} (\vec{r}) = - \vec{\\abla} V (\vec{r}) = - \vec{\\abla} k](https://img.qammunity.org/2020/formulas/physics/college/z5pkhwismeeo1rr80iw66o3t7414o1xxap.png)
![\vec{E} (\vec{r}) = - ((\partial)/(\partial x) , (\partial)/(\partial y ) , (\partial)/(\partial z)) k](https://img.qammunity.org/2020/formulas/physics/college/ha29xbua9o7mtlrsvfu1l2lf1wd6vnspr7.png)
![\vec{E} (\vec{r}) = - ((\partial k)/(\partial x) , (\partial k)/(\partial y ) , (\partial k)/(\partial z))](https://img.qammunity.org/2020/formulas/physics/college/cz2kogfzfqmp7fyqyrrvqwqu8al632bpzw.png)
![\vec{E} (\vec{r}) = - (0,0,0)](https://img.qammunity.org/2020/formulas/physics/college/rae47jvr6y98qv7zcvrd2470idm13n6v6d.png)
This is not the only case in which we would find an zero electric field, as, any scalar field with gradient zero will give an zero electric field. For example:
![V(\vec{r})= (x+2)^2 (y+4)^3 (z+5)^4](https://img.qammunity.org/2020/formulas/physics/college/nai6w7rhog32rx2oaqkmdjm1vea5o36qh7.png)
give an electric field of zero at point (0,0,0)