Question

A 12.5μF capacitor is connected to a power supply that keeps a constant potential difference of 24.0 V across the plates. A piece of material having a dielectric constant of 3.75 is placed between the plates, completely filling the space between them.
a.) How much energy is stored in the capacitor before the dielectric is inserted?
b.) How much energy is stored in the capacitor after the dielectric is inserted?
c.) By how much did the energy change during the insertion? Did it increase or decrease?
d.) Explain why inserting the dielectric (or equivalently exchanging air with the material) causes a change in the stored energy of the capacitor.

Answer 1

Concepts and reason

The main concept used is Energy of capacitor and capacitance of the capacitor.

Initially, calculate the capacitance by using the expression for the capacitance after introducing the dielectric. Later calculate the energy of the capacitor before the slab using the expression of the energy stored in the capacitor. Finally used the change in energy formula to calculate the change in energy before and after introducing the slab.

Fundamentals

The expression for energy stored in a capacitor is,

$U = \frac{1}{2}C{V^2}$

Here, $V$ is the potential difference and $C$ is the capacitance.

(a)

Calculate the energy.

Given capacitance is $12.5{\rm{ \mu F}}$ and potential difference is $24.0{\rm{ V}}$ .

The expression for energy stored in a capacitor is,

$U = \frac{1}{2}C{V^2}$

Substitute $12.5{\rm{ \mu F}}$ for $C$ and $24.0{\rm{ V}}$ for $V$ in above equation.

$\begin{array}{c}\\U = \frac{1}{2}\left( {12.5{\rm{ \mu F}}\left( {\frac{{1{\rm{ F}}}}{{{{10}^6}\,{\rm{\mu F}}}}} \right)} \right){\left( {24.0{\rm{ V}}} \right)^2}\\\\ = 3.6{\rm{ mJ}}\\\end{array}$

(b)

Dielectric constant is given $3.75$ .

New capacitance after the dielectric constant is introduced is,

$C' = \kappa C$

Substitute $12.5{\rm{ \mu F}}$ for $C$ in above equation.

$\begin{array}{c}\\C' = 3.75\left( {12.5{\rm{ \mu F}}} \right)\\\\ = 46.8\;{\rm{\mu F}}\\\end{array}$

New capacitance after the dielectric constant is introduced is $46.8\;{\rm{\mu F}}$ .

The expression for energy stored in a capacitor when dielectric is inserted,

${U_1} = \frac{1}{2}C'{V^2}$

Substitute $46.8\;{\rm{\mu F}}$ for $C'$ and $24.0{\rm{ V}}$ for $V$ in above equation.

$\begin{array}{c}\\{U_1} = \frac{1}{2}\left( {46.8\;{\rm{\mu F}}\left( {\frac{{1{\rm{ F}}}}{{{{10}^6}{\rm{\mu F}}}}} \right)} \right){\left( {24.0{\rm{ V}}} \right)^2}\\\\ = 13.5\;{\rm{mJ}}\\\end{array}$

(c)

The expression for change in energy is,

$\Delta U = {U_1} - U$

Here, ${U_1}$ is the energy stored in the capacitor when dielectric is inserted and $U$ is the energy stored in the capacitor before dielectric is inserted.

Substitute $13.5\;{\rm{mJ}}$ for ${U_1}$ and $3.6{\rm{ mJ}}$ for $U$ in equation of change in energy.

$\begin{array}{c}\\\Delta U = 13.5\;{\rm{mJ}} - 3.6{\rm{ mJ}}\\\\ = 9.9{\rm{ mJ}}\\\end{array}$

Energy stored in capacitor before the dielectric is inserted is $3.6{\rm{ mJ}}$ and when dielectric is inserted energy stored in capacitor is $13.5\;{\rm{mJ}}$ . There is an increase in energy.

[d]

Before the dielectric is inserted, the space between the plates is apparently filled with air.

When dielectric is, inserted capacitance is increased by $\kappa$ factor. Dielectric material between two plates of capacitor get ionized, ionization lead more charge storage inside dielectric. Energy stored in the capacitor is proportional to its capacitance, and the capacitance increases while introducing a dielectric. If the capacitor weren't connected to a battery, the energy would have decreased.

Ans: Part a

$3.6{\rm{ mJ}}$ energy stored in capacitor before the dielectric is inserted.