Find the capacitance of the capacitor 
Given: Consider a capacitor connected to a battery of voltage V . Let the capacitor have an area A, and a distance L between the plates. Assume that the
  capacitor has a layer of linear dielectric (of dielectric constant κ, so that
  ε = κε0) of thickness L/2 on the lower plate.


I found the capacitance to be $C=\frac{2\kappa\epsilon_0 A}{L}$. Was I correct about the separation distance? In more common examples of a capacitor, they claim that the plates are flat with nothing in between. In this case, the linear dielectric is as tall as half the total distance between the plates. Am I correct in using the remaining free space, namely $L/2$, as the separation distance in the equation? For reference, the equation is:
$$C=\frac{\kappa \epsilon_0 A}{d}$$
 A: Instant check that shows your answer is wrong: set $\kappa = 1$ as if there is no dielectric, and you don't recover the parallel plate formula for a gap of $L$.
How to solve this problem:
Let the voltage across the capacitor be $V$ and the voltages across the top and bottom be $V_t$ and $V_b$ respectively, then $$V=V_t+V_b,$$ which implies that $$1/C=1/C_t+1/C_b$$ since $V=Q/C$.
Thus essentially this system consists of two capacitors in series. One capacitor is the top gap, and one is the bottom part. Use the formula for plate capacitors twice, with the separation being $L/2$ in both cases, then get the total capacitance by using the formula for two capacitors in series. Some algebra is involved.
You can check the answer you get, again, by setting $\kappa=1$, and seeing if you recover the usual plate formula.
A: The $\mathbf{D}$ field between the plates is (as usual, neglecting fringing effects) uniform and normal to the plane of the plates (assume in the z direction for simplicity) and is given by
$$\mathbf{D} = \frac{Q}{A}\hat{\mathbf{z}}$$
where the plates have charge $Q$ and $-Q$ respectively.  The electric field (between the plates) within the dielectric is
$$\mathbf{E_\kappa} = \frac{\mathbf{D}}{\kappa \epsilon_0}$$
and the electric field (between the plates) in air is
$$\mathbf{E} = \frac{\mathbf{D}}{\epsilon_0}$$
Thus, the potential difference between the plates is
$$V = \left(\frac{L}{2}\mathbf{E_\kappa} +  \frac{L}{2}\mathbf{E}\right)\cdot\hat{\mathbf{z}} = \left(\frac{L}{2}\frac{\mathbf{D}}{\kappa \epsilon_0} +  \frac{L}{2}\frac{\mathbf{D}}{\epsilon_0}\right)\cdot\hat{\mathbf{z}}  = \left(\frac{L}{2}\frac{Q}{A\kappa\epsilon_0} + \frac{L}{2}\frac{Q}{A\epsilon_0} \right)$$
The capacitance is then
$$C = \frac{Q}{V} = \frac{1}{\frac{L/2}{A\kappa\epsilon_0} + \frac{L/2}{A\epsilon_0}} = \frac{1}{\frac{1}{C_1}+ \frac{1}{C_2}}$$
which is the formula for series connected capacitors.
