Charged Hemisphere The potential of a hemisphere  at the centre with constant surface charge density $\sigma $ is given by $\frac { \sigma R }{ 2\epsilon  } $ where $R$ is the radius of the hemisphere. The magnitude of the electric field for the same configuration is given by $\frac { -\sigma  }{ 4\epsilon  } $.
I've got the potential by doing a surface integral and the field by integrating rings that make the hemisphere. 
But why can't we use $E=-\nabla V$ arriving at $\frac { -\sigma  }{ 2\epsilon  } $ for the field.
Thanks. 
 A: Given you have evaluated the potential as number, you cannot take the derivative of this number w/r to anything.  
To proceed with the gradient, you need to find $V$ as a function of the position (by symmetry you can restrict to the position on the axis), i.e. $V(z)$ and then $\vec E=-\hat z \frac{\partial V}{\partial z}$ since by symmetry the field cannot have components along any other axis.

Edit: 
In your specific example, you can argue by symmetry that the field $\vec E$ for a point on the axis of symmetry will be along $\boldsymbol{\hat z}$.  Hence, you compute the potential for any point $V(z)$ on the symmetry axis (rather than just at the center of the hemisphere), you can compute $E_z=-\frac{\partial V}{\partial z}$.
With reference to the geometry below (where the radius of the sphere is $a$ rather than your $R$)

the expression for the potential at a point $P$ located at $(0,0,z)$  on the symmetry axis with $z>0$ is
$$
V(z) = \frac{Q}{4\pi\epsilon_0}\frac{(a+z -\sqrt{a^2+z^2})}{az}\, . \tag{1}
$$
so you can easily recover the field from that. 
Obtaining (1) is reasonably straightforward although there is a possibly tricky integral which you might have to look up. 
It might also be useful to know that, for $z$ small but positive, a series expansion of $\frac{(a+z -\sqrt{a^2+z^2})}{az}$ shows that the potential remains finite at $z=0$ and in fact has a part linear in $z$, so the field will also be finite at $z=0$.
[Please note that I copied (1) from an old textbook: I do not guarantee it is correct.  You will have to check this yourself but given the behaviour for small $z$ Eq.(1) seems like a reasonable expression.]
