A parallel plate capacitor is maintained at a certain potential difference. When a dielectric slab of thickness $3 \mathrm{~mm}$ is introduced between the plates, the plate separation had to be increased by $2 \mathrm{~mm}$ in order to maintain the same potential difference between the plates. The…

Question

A parallel plate capacitor is maintained at a certain potential difference. When a dielectric slab of thickness $3 \mathrm{~mm}$ is introduced between the plates, the plate separation had to be increased by $2 \mathrm{~mm}$ in order to maintain the same potential difference between the plates. The dielectric constant of the slab is

Aicharya · Accepted Answer

$V^{\prime}=\frac{Q}{C^{1}}=\frac{Q\left[d^{\prime}-t\left(1-\frac{1}{K}\right)\right]}{\varepsilon_{0} A}, \quad V=\frac{Q}{C}=\frac{Q d}{\varepsilon_{0} A}$
Now, $V^{\prime}=V \Rightarrow d=d^{\prime}-t\left(1-\frac{1}{K}\right)$
$d^{\prime}-d=2 \mathrm{~mm}, t=3 \mathrm{~mm} \quad \Rightarrow \quad 2=3\left(1-\frac{1}{K}\right) \Rightarrow K=3$ ^

A parallel plate capacitor is maintained at a certain potential difference. When a dielectric slab of…

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