A parallel plate capacitor having a separation between the plates d, plate area A and material with dielectric constant K has capacitance $$C_0$$. Now one-third of the material is replaced by another material with dielectric constant 2K, so that effectively there are two capacitors one with area $$\frac{1}{3}$$ A, dielectric constant 2K and another with area $$\frac{2}{3}$$ A and dielectric constant K. If the capacitance of this new capacitor is C then $$\frac{C}{C_0}$$ is

First, recall the formula for the capacitance of a parallel plate capacitor with a dielectric material. The capacitance is given by $$ C = \frac{K \epsilon_0 A}{d} $$, where $$ K $$ is the dielectric constant, $$ \epsilon_0 $$ is the permittivity of free space, $$ A $$ is the plate area, and $$ d $$ is the separation between the plates.

For the initial capacitor, the dielectric constant is $$ K $$, plate area is $$ A $$, and separation is $$ d $$. So, the initial capacitance $$ C_0 $$ is:

$$ C_0 = \frac{K \epsilon_0 A}{d} $$

Now, one-third of the dielectric material is replaced by another material with dielectric constant $$ 2K $$. As described, this effectively creates two capacitors in parallel: one with area $$ \frac{1}{3}A $$ and dielectric constant $$ 2K $$, and another with area $$ \frac{2}{3}A $$ and dielectric constant $$ K $$. Since they share the same plate separation $$ d $$ and are connected across the same plates, they are in parallel.

For capacitors in parallel, the total capacitance $$ C $$ is the sum of the individual capacitances. So, we need to find the capacitance of each part and add them together.

Start with the first capacitor, which has area $$ A_1 = \frac{1}{3}A $$ and dielectric constant $$ K_1 = 2K $$. Its capacitance $$ C_1 $$ is:

$$ C_1 = \frac{K_1 \epsilon_0 A_1}{d} = \frac{(2K) \epsilon_0 \left(\frac{1}{3}A\right)}{d} $$

Simplify the expression:

$$ C_1 = \frac{2K \epsilon_0 \cdot \frac{1}{3}A}{d} = \frac{2K \epsilon_0 A}{3d} $$

Notice that $$ \frac{K \epsilon_0 A}{d} $$ is exactly $$ C_0 $$, from the initial capacitance formula. So, we can write $$ C_1 $$ in terms of $$ C_0 $$:

$$ C_1 = \frac{2}{3} \times \frac{K \epsilon_0 A}{d} = \frac{2}{3} C_0 $$

Now, for the second capacitor, it has area $$ A_2 = \frac{2}{3}A $$ and dielectric constant $$ K_2 = K $$. Its capacitance $$ C_2 $$ is:

$$ C_2 = \frac{K_2 \epsilon_0 A_2}{d} = \frac{K \epsilon_0 \left(\frac{2}{3}A\right)}{d} $$

Simplify:

$$ C_2 = \frac{K \epsilon_0 \cdot \frac{2}{3}A}{d} = \frac{2K \epsilon_0 A}{3d} $$

Again, since $$ \frac{K \epsilon_0 A}{d} = C_0 $$, we have:

$$ C_2 = \frac{2}{3} \times \frac{K \epsilon_0 A}{d} = \frac{2}{3} C_0 $$

Now, the total capacitance $$ C $$ of the new capacitor is the sum of $$ C_1 $$ and $$ C_2 $$ because they are in parallel:

$$ C = C_1 + C_2 = \frac{2}{3} C_0 + \frac{2}{3} C_0 = \frac{4}{3} C_0 $$

Therefore, the ratio $$ \frac{C}{C_0} $$ is:

$$ \frac{C}{C_0} = \frac{\frac{4}{3} C_0}{C_0} = \frac{4}{3} $$

Hence, the ratio $$ \frac{C}{C_0} $$ is $$ \frac{4}{3} $$. Comparing with the options, this corresponds to Option B.

So, the answer is $$ \frac{4}{3} $$.