Question

An air-filled parallel-plate capacitor has a capacitance of 1.4 pF. The separation of the plates is doubled and wax is inserted between them. The new capacitance is 2.1 pF. Find the dielectric constant of the wax.  

Answer 1

Concepts and reason

The concepts of capacitance and the dielectric constant are used in this problem. The dielectric constant of the wax can be calculated using the formula of capacitance in terms of dielectric constant.

Fundamentals

Capacitance:

The capacitance is the property of a capacitor which measures its ability to store electric charge. The formula for capacitance is:

$C = \frac{{k{\varepsilon _{\rm{o}}}A}}{d}$

Here, $k$ is the dielectric constant, ${\varepsilon _{\rm{o}}}$ is the permittivity of free space, $A$ is the surface area of the plates and $d$ is the distance between the parallel plates.

The formula of capacitance is:

$C = \frac{{k{\varepsilon _{\rm{o}}}A}}{d}$

When air is filled between plates, the capacitance is:

${C_{\rm{a}}} = \frac{{{k_{\rm{a}}}{\varepsilon _{\rm{o}}}A}}{{{d_{\rm{a}}}}}$

Here, ${k_{\rm{a}}}$ is the dielectric constant of air, ${d_{\rm{a}}}$ is the distance between two plates when air is filled between them.

When wax is filled between plates, the capacitance is:

${C_{\rm{w}}} = \frac{{{k_{\rm{w}}}{\varepsilon _{\rm{o}}}A}}{{{d_{\rm{w}}}}}$

Here, ${k_{\rm{w}}}$ is the dielectric constant of wax, ${d_{\rm{w}}}$ is the distance between two plates when wax is filled between them.

Substitute $2{d_{\rm{a}}}$ for ${d_{\rm{w}}}$ .

${C_{\rm{w}}} = \frac{{{k_{\rm{w}}}{\varepsilon _{\rm{o}}}A}}{{2{d_{\rm{a}}}}}$

Taking the ratio of both capacitances:

$\begin{array}{c}\\\frac{{{C_{\rm{a}}}}}{{{C_{\rm{w}}}}} = \frac{{\frac{{{k_{\rm{a}}}{\varepsilon _{\rm{o}}}A}}{{{d_{\rm{a}}}}}}}{{\frac{{{k_{\rm{w}}}{\varepsilon _{\rm{o}}}A}}{{2{d_{\rm{a}}}}}}}\\\\\frac{{{C_{\rm{a}}}}}{{{C_{\rm{w}}}}} = \frac{{2{k_{\rm{a}}}}}{{{k_{\rm{w}}}}}\\\\{k_{\rm{w}}} = \frac{{2{k_{\rm{a}}}{C_{\rm{w}}}}}{{{C_{\rm{a}}}}}\\\end{array}$

The dielectric constant of wax is:

${k_{\rm{w}}} = \frac{{2{k_{\rm{a}}}{C_{\rm{w}}}}}{{{C_{\rm{a}}}}}$

Substitute $1$ for ${k_{\rm{a}}}$ , $2.1{\rm{ pF}}$ for ${C_{\rm{w}}}$ and $1.4{\rm{ pF}}$ for ${C_{\rm{a}}}$ .

$\begin{array}{c}\\{k_{\rm{w}}} = \frac{{2\left( 1 \right)\left( {2.1{\rm{ pF}}} \right)}}{{\left( {1.4{\rm{ pF}}} \right)}}\\\\ = 2\left( {\frac{3}{2}} \right)\\\\ = {\bf{3}}\\\end{array}$

Ans:

The dielectric constant of the wax is ${\bf{3}}$ .