# Why the euclid distance between the plates of the capacitor has been doubled to find out the capacitance?

I may have been asking of an easy problem.

The problem requires to find out the capacitance.

The ideal capacitor with area of $$\S\$$ exists and the euclid distance between the plates is $$\d\$$.

The dielectric with permittivity of $$\\epsilon\$$ has been filled in the half right side of the space between the plates and the remaining half left side has been filled by vacuum(of course the permittivity of it is $$\\epsilon_0\$$).

The textbook solution says that the capacitor of the problem is equivalent to the connected 2 capacitors in parallel.

One of the plates of the parallel capacitors has been filled with the dielectric and the other has been filled with vacuum.

$$\C_0:=\frac{\epsilon_0 S}{2d}\$$(capacitance of the capacitor with vacuum of parallel)

$$\C_1:=\frac{\epsilon S}{2d}\$$(capacitance of the capacitor of the dielectric of parallel)

And my doubt has came from here.

Why the each denominator has $$\2\$$?

I thought that the each denominator must be $$\d\$$.

And the capacitance of the answer $$\C\$$ is given by $$\C=C_0+C_1\$$.

Can anyone tell me the hints so that I can resolve the doubt on my own.

$$\\$$

If we wrote it as $$C_0 := \frac 1 2 \frac {\epsilon _0 S} {d}$$ $$C_1 := \frac 1 2 \frac {\epsilon S} {d}$$
would it make more sense? Can you see where $$\ \frac 1 2 \$$ would come in?