Q149 / 160 · TS EAMCET 2004 · Physics

The plates of a parallel plate capacitor are charged upto 200 volts. A dielectirc slab of thickness $4 \mathrm{~mm}$ is inserted between its plates. Then, to maintain the same potential difference between the plates of the capacitor, the distance between the plates is increased by $3.2 \mathrm{~mm}$. The dielectric constant of the dielectric slab is

  • A
    $1$
  • B
    $4$
  • Correct answer
    C
    $5$
  • D
    $6$

Solution

$$ \begin{aligned} & \frac{\varepsilon_0 A}{d}=\frac{\varepsilon_0 A}{d^{\prime}-t+\frac{t}{K}} \\ & \Rightarrow \quad d=d^{\prime}-t+\frac{t}{K} \Rightarrow d^{\prime}-d=t\left(1-\frac{1}{K}\right) \end{aligned} $$ Here, $d^{\prime}-d=3.2 \mathrm{~mm}, t=4 \mathrm{~mm}$ $$ \begin{aligned} & \therefore \quad 3.2=4\left(1-\frac{1}{K}\right) 1-\frac{1}{K} \\ & \Rightarrow \quad \frac{3.2}{4}=1-\frac{1}{K} \Rightarrow 1-\frac{1}{K}=\frac{4}{5} \\ & \therefore \quad K=5 \\ & \end{aligned} $$

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

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