Answer:
Explanation:
In electrostatics, the electric field [tex]\vec{E}[/tex] is related to the gradient of the electric potential V with :
[tex]\vec{E} (\vec{r}) = - \vec{\nabla} V (\vec{r})[/tex]
This means that for constant electric potential the electric field must be zero:
[tex]V(\vec{r}) = k[/tex]
[tex]\vec{E} (\vec{r}) = - \vec{\nabla} V (\vec{r}) = - \vec{\nabla} k [/tex]
[tex]\vec{E} (\vec{r}) = - (\frac{\partial}{\partial x} , \frac{\partial}{\partial y } , \frac{\partial}{\partial z}) k [/tex]
[tex]\vec{E} (\vec{r}) = - (\frac{\partial k}{\partial x} , \frac{\partial k}{\partial y } , \frac{\partial k}{\partial z}) [/tex]
[tex]\vec{E} (\vec{r}) = - (0,0,0) [/tex]
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:
[tex]V(\vec{r})= (x+2)^2 (y+4)^3 (z+5)^4[/tex]
give an electric field of zero at point (0,0,0)
