The plates of a parallel plate capacitor are charged upto $100 \mathrm{~V}$. A $2 \mathrm{~mm}$ thick insulator sheet is inserted between the plates. Then to maintain the same potential difference, the distance between the plates is increased by $1.6 \mathrm{~mm}$. The dielectric constant of the…

Question

The plates of a parallel plate capacitor are charged upto $100 \mathrm{~V}$. A $2 \mathrm{~mm}$ thick insulator sheet is inserted between the plates. Then to maintain the same potential difference, the distance between the plates is increased by $1.6 \mathrm{~mm}$. The dielectric constant of the insulator is

Aicharya · Accepted Answer

Potential of capacitor before inserting insulator, $ V=\frac{\sigma d}{\varepsilon_0} $ where, $\sigma=$ surface charge density, $d=$ distance between plates

and $\varepsilon_0=$ permittivity of vacuum. Given, $\frac{\sigma d}{\varepsilon_0}=100$ Now, potential after inserting $2 \mathrm{~mm}$ of insulator. Let its dielectric constant is $k$. $ \begin{aligned} & V_1=E_1 \cdot 2+E_2(d-2) \\ & V_1=\frac{\sigma}{\varepsilon_0 k} 2+\frac{\sigma}{\varepsilon_0}(d-2) \Rightarrow \quad V_1=\frac{\sigma}{\varepsilon_0}\left(\frac{2}{k}+d-2\right) \end{aligned} $ To maintain potential difference, same distance is increased by $1.6 \mathrm{~mm}$ So, extra potential difference, $V_2=\frac{\sigma}{\varepsilon_0}(1.6)$ Clearly, $V_1+V_2=V$ $ \begin{array}{lc} & \frac{\sigma}{\varepsilon_0}\left(\frac{2}{k}+d-2\right)+\frac{\sigma}{\varepsilon_0} 1.6=\frac{\sigma d}{\varepsilon_0} \\ \Rightarrow & \frac{\sigma d}{\varepsilon_0}-\frac{2 \sigma}{\varepsilon_0 k}-\frac{\sigma}{\varepsilon_0}(d-2)=\frac{\sigma}{\varepsilon_0} 1.6 \\ \Rightarrow & \frac{2 \sigma}{\varepsilon_0}-\frac{2 \sigma}{\varepsilon_0 k}=\frac{\sigma}{\varepsilon_0} 1.6 \Rightarrow 2-\frac{2}{k}=1.6 \\ \Rightarrow & 2 k-2=1.6 k \\ \Rightarrow & \quad k=5 \end{array} $

The plates of a parallel plate capacitor are charged upto $100 \mathrm{~V}$. A $2 \mathrm{~mm}$ thick…

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