A parallel plate capacitor with plate area A and plate separation d is filled with a dielectric material of dielectric constant K=4 . The thickness of the dielectric material is x , where x \lt d . Let C_1 and C_2 be the capacitance of the system for \chi=\frac13 d and \mathcalX=\frac2 d3 , respectively. If C_1=2 \mu F the value of C_2 is \underline\hspace2cm \mu F

Capacitor PYQ — JEE Main 2023 · 6 Apr · Shift 1 · Q69 with Solution

View written solutionFree

Correct answer: 3

A dielectric slab of thickness $x$ fills the full plate area $A$ between plates separated by $d$ .

So the space between the plates has:

These two regions are along the field direction, so they behave like two capacitors in series.

For series layers,

\frac{1}{C}=\frac{x}{K\varepsilon_0 A}+\frac{d-x}{\varepsilon_0 A}

Thus,

\frac{1}{C}=\frac{1}{\varepsilon_0 A}\left(\frac{x}{K}+d-x\right)

Hence,

C=\frac{\varepsilon_0 A}{d-x+\frac{x}{K}}

Given $K=4$ ,

C=\frac{\varepsilon_0 A}{d-x+\frac{x}{4}} =\frac{\varepsilon_0 A}{d-\frac{3x}{4}}

C_1=\frac{\varepsilon_0 A}{d-\frac{3}{4}\cdot \frac{d}{3}} =\frac{\varepsilon_0 A}{d-\frac{d}{4}} =\frac{\varepsilon_0 A}{\frac{3d}{4}} =\frac{4\varepsilon_0 A}{3d}

Given,

C_1=2\,\mu F

So,

\frac{4\varepsilon_0 A}{3d}=2

This gives

\frac{\varepsilon_0 A}{d}=\frac{3}{2}\,\mu F

C_2=\frac{\varepsilon_0 A}{d-\frac{3}{4}\cdot \frac{2d}{3}} =\frac{\varepsilon_0 A}{d-\frac{d}{2}} =\frac{\varepsilon_0 A}{\frac{d}{2}} =2\frac{\varepsilon_0 A}{d}

Using

\frac{\varepsilon_0 A}{d}=\frac{3}{2}\,\mu F

we get

C_2=2\times \frac{3}{2}=3\,\mu F

\boxed{3}