As shown in the figure, two parallel plate capacitors having equal plate area of 200 cm$$^2$$ are joined in such a way that $$a \neq b$$. The equivalent capacitance of the combination is $$x\varepsilon_0 F$$. The value of $$x$$ is ______.

So

$$\frac{1}{C_{eq}}=\frac{1}{C_1}+\frac{1}{C_2}$$

with

$$C_1=\frac{\varepsilon_0A}{a}$$

$$C_2=\frac{\varepsilon_0A}{b}$$

Thus

$$\frac{1}{C_{eq}}=\frac{a}{\varepsilon_0A}+\frac{b}{\varepsilon_0A}$$

$$\frac{1}{C_{eq}}=\frac{a+b}{\varepsilon_0A}$$

So

$$C_{eq}=\frac{\varepsilon_0A}{a+b}$$

Now from geometry,

$$a+b=d-c$$

Given

d=5 mm,c=1 mm so

a+b=4 mm=$$4\times\ 10^{-3}\ m$$ 

Area:

A=200 $$cm^2$$

$$=2\times10^{-2}m$$

$$=2\times10^{-2}m^2$$

Hence

$$C_{eq}=\frac{\varepsilon_0(2\times10^{-2})}{4\times10^{-3}}$$

$$=5\varepsilon_0$$

Comparing with

$$C_{eq}=x\varepsilon_0$$

we get

x=5