If the electrostatic potential is zero, why doesn't the electric field have to be zero? I thought the relation between the electrostatic field $\vec E$ and the electrostatic potential $V$ is as follows:
$$\vec E = - \nabla V$$
Thus, when $V$ is zero, $\vec E$ is also zero.
 A: It depends on what you mean when you say $V=0$. In the context of the equation:
$$\vec{E}=-\nabla V$$
which holds specifically in electrostatics $V$ is a scalar field, meaning that it is actually a function which assigns every point in space a scalar value. $\vec{E}$ is a vector field, which assigns a vector to every point in space. Thus, both the electric field and the potential are dependent upon position. This can be shown more explicitly as:
$$\vec{E}(\textbf{r})=-\nabla V(\textbf{r})$$
where $\textbf{r}$ is a position vector. Now, if $V(\textbf{r})=0$ for all $\textbf{r}$ then certainly the gradient is also zero everywhere, and thus, the electric field is zero everywhere.
On the other hand $V(\textbf{r})$ may equal zero for only some $\textbf{r}$. For example, at the point $P$ midway between two point charges, one with charge $+q$ and the other with charge $-q$ the potential is zero, assuming infinity as the reference point. However, if you move even slightly away from this point, the potential is non zero. The fact that the potential is changing at point $P$ indicates that the gradient at this point is non zero. Thus, the electric field at $P$ is non zero, even though the potential itself is zero at $P$. 
Note that this is true for electrostatics, but, as Sebastian mentions in a comment below, it is incomplete in the context of electrodynamics. This is simply because the expression you point to relating the electric field to the potential only holds for electrostatics. For treatment of the more general case, please see Sebastian's and Alexander's fine answers.
A: This is up to a gauge you're using.
The most general expression is -
$$ \vec{E}=-\nabla\phi-\frac{\partial\vec{A}}{\partial t} $$
$$ \vec{B}=\nabla \times \vec{A} $$
so it's possible to nullify $\phi $ and still get any electromagnetic field you want.
A: The relationship between electrical field and scalar potential you give only holds
in the static case (or after an gauge transformation).
The full formula for the case of time dependent fields is:
$$ \vec E = - \nabla V - \partial_t \vec A $$
The magnetic field then is given by:
$$ \vec B = \nabla \times \vec A. $$
A: $$\vec{E}=-\vec{\nabla}V$$
means: the electric field is the derivative (3 dimensional) of V. So wether $V = 0$ or not, doesn't matter. It's wether V changes or not, that defines the electric field.
An analogy to make things clear: analogy for potential = height, analogy for electric field= 'how does the ball roll?'. To see 'How does the ball roll?', you don't need to know the absolute height, only wether there is a slope (a difference in height): an equal slope has the same effect on the ball in the Netherlands as in Nepal.  
A: Electric field is vector so there is a possibility for the electric field to be zero at a point but it isn't the same with the electric potential it is a scalar ie the net potential is the algebraic sum of individual potentials so it is not necessary for potential to be zero if field is zero and vice versa hope u understood
