Question

An isolated parallel plate capacitor is charged upto a certain potential difference. When a $3\,{\text{mm}}$ thick slab is introduced between the plates then in order to maintain the same potential difference, the distance between the plates is increased by $2.4\,{\text{mm}}$. Find the dielectric constant of the slab. (assume charge remains constant)

Answer 1

Use the expressions for the capacitance of a parallel plate capacitor and the capacitance of a parallel plate capacitor when a dielectric slab is placed between its plates. Equate these equations of capacitance to determine the dielectric constant of the slab.

Formulae used:

The expression for the capacitance $C$ of a parallel plate capacitor is
$C = \dfrac{{A{\varepsilon _0}}}{d}$ …… (1)

Here, $A$ is the area of the plate, ${\varepsilon _0}$ is the permittivity of the free space and $d$ is the distance between the plates of the capacitor.

The expression for the capacitance $C$ of a parallel plate capacitor when a dielectric slab is placed between its plates is
$C = \dfrac{{A{\varepsilon _0}}}{{d' - t\left( {1 - \dfrac{1}{K}} \right)}}$ …… (2)

Here, $A$ is the area of the plate, ${\varepsilon _0}$ is the permittivity of the free space, $d'$ is the distance between the plates of capacitor, $t$ is the thickness of the slab and $K$ is the dielectric constant.

The formula for the capacitance $C$ is given by
$C = \dfrac{q}{v}$ …… (3)

Here, $q$ is the charge stored on the capacitor and $v$ is the potential difference between the plates of the capacitor.

Complete step by step answer: An isolated parallel plate capacitor is charged upto a certain potential difference.

The distance $d'$ between the plates of capacitor increases by $2.4\,{\text{mm}}$ when a slab of thickness $3\,{\text{mm}}$ is placed between the plates of the capacitors.
$d' = \left( {d + 2.4\,{\text{mm}}} \right)$

Here, $d$ is the initial distance between the plates of the capacitor.

The capacitance ${C_i}$ of the capacitor before placing slab between its plates is
${C_i} = \dfrac{{A{\varepsilon _0}}}{d}$

The capacitance ${C_f}$ of the capacitor after placing slab between its plates is
${C_f} = \dfrac{{A{\varepsilon _0}}}{{d' - t\left( {1 - \dfrac{1}{K}} \right)}}$

The charge stored by the capacitor and potential difference remains the same before and after placing the slab between the plates of the capacitor.

Hence, from equation (3), the capacitances ${C_i}$ and ${C_f}$ of the capacitor also remain the same.
${C_i} = {C_f}$

Substitute $\dfrac{{A{\varepsilon _0}}}{d}$ for ${C_i}$ and $\dfrac{{A{\varepsilon _0}}}{{d' - t\left( {1 - \dfrac{1}{K}} \right)}}$ for ${C_f}$ in the above equation.
$\dfrac{{A{\varepsilon _0}}}{d} = \dfrac{{A{\varepsilon _0}}}{{d' - t\left( {1 - \dfrac{1}{K}} \right)}}$
$\Rightarrow d = d' - t\left( {1 - \dfrac{1}{K}} \right)$

Substitute $d + 2.4\,{\text{mm}}$ for $d'$ and $3\,{\text{mm}}$ for $t$in the above equation.
$d = \left( {d + 2.4\,{\text{mm}}} \right) - \left( {3\,{\text{mm}}} \right)\left( {1 - \dfrac{1}{K}} \right)$
$\Rightarrow \left( {3\,{\text{mm}}} \right) - \dfrac{{\left( {3\,{\text{mm}}} \right)}}{K} = \left( {2.4\,{\text{mm}}} \right)$
$\Rightarrow \left( {3\,{\text{mm}}} \right) - \left( {2.4\,{\text{mm}}} \right) = \dfrac{{\left( {3\,{\text{mm}}} \right)}}{K}$
$\Rightarrow 0.6\,{\text{mm}} = \dfrac{{\left( {3\,{\text{mm}}} \right)}}{K}$

Rearrange the above equation for the dielectric constant $K$.
$\Rightarrow K = \dfrac{{3\,{\text{mm}}}}{{0.6\,{\text{mm}}}}$
$\Rightarrow K = 5$

Hence, the dielectric constant of the slab is 5.

Note: There is no need to convert the unit of the thickness of the slab and distance between the plates in the SI system of units as the dielectric constant has no unit.