Q86 / 100 · BITSAT 2024 (Memory Based Paper 1) · Physics

A parallel plate capacitor with plate area A and plate separation $d=2 \mathrm{~m}$ has a capacitance of $4 \mu \mathrm{~F}$. The new capacitance of the system if half of the space between them is filled with a dielectric material of dielectric constant $\mathrm{K}=3$ (as shown in figure) will be: Figure

  • A
    $2 \mu \mathrm{~F}$
  • B
    $32 \mu \mathrm{~F}$
  • Correct answer
    C
    $6 \mu \mathrm{~F}$
  • D
    $8 \mu \mathrm{~F}$

Solution

We have, $C_i=\frac{A \epsilon_0}{d}=4 \mu F$ $C_f=\frac{A \epsilon_0}{d-t+\frac{t}{k}}=\frac{A \epsilon_0}{d-\frac{d}{2}+\frac{d}{2 \times 3}}$ $=\frac{A \in_0}{d\left(1-\frac{1}{2}+\frac{1}{6}\right)}$ $=\frac{4 \mu F}{\frac{2}{3}}=6 \mu F$

Hence, the correct answer is option: 'C'.

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