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

Correct answer: B. (4)/(3) C_0= k ∈_0 ~A ~d C= k ∈_0 2 3 ~d + 2 k ∈_0 ~A 3 ~d =(4)/(3) k ∈_0 ~A ~d C C_0 = (4)/(3) k ∈_0 ~A ~d k ∈_0 ~A ~d =(4)/(3)

JEE Main 201325 AprilOnline

JEE Main 2013 · 25 April · Online · Physics · Capacitance

A parallel plate capacitor having a separation between the plates d...

A parallel plate capacitor having a separation between the plates

\mathrm{d}

, plate area

\mathrm{A}

and material with dielectric constant

\mathrm{K}

has capacitance

\mathrm{C}_0

. Now one-third of the material is replaced by another material with dielectric constant

2 \mathrm{~K}

, so that effectively there are two capacitors one with area

\frac{1}{3} \mathrm{~A}

, dielectric constant

2 \mathrm{~K}

and another with area

\frac{2}{3} \mathrm{~A}

and dielectric constant

\mathrm{K}

. If the capacitance of this new capacitor is

\mathrm{C}

then

\frac{\mathrm{C}}{\mathrm{C}_0}

B. $\frac{4}{3}$

\mathrm{C}_0=\frac{\mathrm{k} \in_0 \mathrm{~A}}{\mathrm{~d}}

\mathrm{C}=\frac{\mathrm{k} \in_0 2}{3 \mathrm{~d}}+\frac{2 \mathrm{k} \in_0 \mathrm{~A}}{3 \mathrm{~d}}=\frac{4}{3} \frac{\mathrm{k} \in_0 \mathrm{~A}}{\mathrm{~d}}

\therefore \frac{\mathrm{C}}{\mathrm{C}_0}=\frac{\frac{4}{3} \frac{\mathrm{k} \in_0 \mathrm{~A}}{\mathrm{~d}}}{\frac{\mathrm{k} \in_0 \mathrm{~A}}{\mathrm{~d}}}=\frac{4}{3}