A parallel plate capacitor has capacitance C, when there is vacuum within the parallel plates. A sheet having thickness $$\left(\frac{1}{3}\right)^{rd}$$ of the separation between the plates and relative permittivity K is introduced between the plates. The new capacitance of the system is:

A parallel plate capacitor of capacitance $$C$$ in vacuum, with plate area $$A$$ and plate separation $$d$$, has a dielectric sheet of thickness $$\frac{d}{3}$$ and relative permittivity $$K$$ introduced between the plates. To determine the new capacitance $$C'$$, we first recall the original expression for a vacuum-filled parallel plate capacitor: $$C = \frac{\epsilon_0 A}{d}$$.

After inserting the dielectric, the system can be modeled as two capacitors in series. The region containing the dielectric has thickness $$t = \frac{d}{3}$$ and permittivity $$K\epsilon_0$$, leading to $$C_1 = \frac{K\epsilon_0 A}{d/3} = \frac{3K\epsilon_0 A}{d}\,. $$ The remaining vacuum region has thickness $$\frac{2d}{3}$$ and permittivity $$\epsilon_0$$, giving $$C_2 = \frac{\epsilon_0 A}{2d/3} = \frac{3\epsilon_0 A}{2d}\,. $$

For capacitors in series, $$\frac{1}{C'} = \frac{1}{C_1} + \frac{1}{C_2}\,. $$ Substituting the expressions for $$C_1$$ and $$C_2$$ yields $$\frac{1}{C'} = \frac{d}{3K\epsilon_0 A} + \frac{2d}{3\epsilon_0 A} = \frac{d}{3\epsilon_0 A}\Bigl(\frac{1}{K} + 2\Bigr) = \frac{d}{3\epsilon_0 A}\,\frac{1 + 2K}{K}\,. $$

Inverting this result gives $$C' = \frac{3K\epsilon_0 A}{d(2K + 1)}\,. $$ Since $$C = \frac{\epsilon_0 A}{d}\,, $$ it follows that $$C' = \frac{3KC}{2K + 1}\,. $$

Thus, the new capacitance is $$\frac{3KC}{2K+1}\,. $$