For changing the capacitance of a given parallel plate capacitor, a dielectric material of dielectric constant K is used, which has the same area as the plates of the capacitor. The thickness of the dielectric slab is 3 \over 4 d, where 'd' is the separation between the plates of parallel plate capacitor. The new capacitance (C') in terms of original capacitance (C0) is given by the following relation :

Capacitor PYQ — JEE Main 2021 · 16 Mar · Shift 1 · Q52 with Solution

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Correct answer: C

For a parallel plate capacitor with plate area $A$ and separation $d$ ,

$C_0 = \frac{\varepsilon_0 A}{d}$

A dielectric slab of dielectric constant $K$ and thickness

$t = \frac{3d}{4}$

is inserted between the plates. Since the slab has the same area as the plates, the system behaves like two capacitors in series along the thickness:

For layered media along the separation direction,

$C' = \frac{\varepsilon_0 A}{\left(d-t\right)+\frac{t}{K}}$

Substitute $t = \frac{3d}{4}$ :

$C' = \frac{\varepsilon_0 A}{\frac{d}{4} + \frac{3d}{4K}}$

Take $\frac{d}{4}$ common from the denominator:

$C' = \frac{\varepsilon_0 A}{\frac{d}{4}\left(1+\frac{3}{K}\right)}$

$C' = \frac{4\varepsilon_0 A}{d\left(1+\frac{3}{K}\right)}$

Now simplify:

$C' = \frac{4\varepsilon_0 A}{d\cdot \frac{K+3}{K}} = \frac{4K\varepsilon_0 A}{d(K+3)}$

Using $C_0 = \frac{\varepsilon_0 A}{d}$ ,

$C' = \frac{4K}{K+3} C_0$

This corresponds to:

$\boxed{C' = \frac{4K}{K+3}C_0}$

So, the correct option is C.