Question

Two oppositely charged but otherwise identical conductingplates of area 2.50 square centimeters are separated by adielectric 1.80 millimeters thick, with a dielectric constant of $K=3.60$ . The resultant electric field in the dielectric is $1.20 \times 10^{6}$ volts per meter.

a) Compute the magnitude of the charge per unit area on the conducting plate.

b) Compute the magnitude of the charge per unit area $sigma_1$ on the surfaces of the dielectric.

c) Find the total electric-field energy stored in the capacitor.

Answer 1

Concepts and reason

Use the concept of Capacitance of a capacitor with dielectric to solve this problem.

The magnitude of the charge per unit area on the conducting plate is calculated using the expression of charge density.

The magnitude of charge per unit area on the surface of dielectric plate is calculated using the charge density of conducting plate.

The total electric - field energy stored in the capacitor is calculated using the expression for energy stored in the capacitor.

Fundamentals

The expression of charge per unit area on the conducting plate is,

$\sigma = k{\varepsilon _0}E$

Here, $k$ is the dielectric constant, $E$ is the magnitude of electric field, ${\varepsilon _0}$ is the permittivity of free space, and $\sigma$ is the charge density.

The expression for magnitude of charge per unit area on the surface of dielectric plate is,

$\sigma ' = \sigma \left( {1 - \frac{1}{k}} \right)$

Here, $\sigma$ is the charge density of conducting plate ${\sigma _1}$ is the charge density on surface of dielectric plate, and k is the dielectric constant.

The expression to calculate the total energy stored in the capacitor is,

$U = \frac{1}{2}k{\varepsilon _0}{E^2}Ad$

Here, $k$ is the dielectric constant, A is the area, d is the thickness of the plates, E is the magnitude of electric field, ${\varepsilon _0}$is the permittivity of free space, and $U$is the energy stored in the capacitor.

(a)

Calculate the magnitude of the charge per unit area on the conducting plate.

The expression of charge per unit area on the conducting plate is,

$\sigma = k{\varepsilon _0}E$

Here, $k$ is the dielectric constant, $E$ is the magnitude of electric field, ${\varepsilon _0}$ is the permittivity of free space, and $\sigma$ is the charge density per unit area.

Substitute 3.60 for k , $8.85 \times {10^{ - 12}}\,{{\rm{C}}^{\rm{2}}}{\rm{/N}} \cdot {{\rm{m}}^{\rm{2}}}$ for ${\varepsilon _0}$, and $1.20 \times {10^6}\,{\rm{V/m}}$for $E$.

$\begin{array}{c}\\\sigma = \left( {3.60} \right)\left( {8.85 \times {{10}^{ - 12}}\,{{\rm{C}}^{\rm{2}}}{\rm{/N}} \cdot {{\rm{m}}^{\rm{2}}}} \right)\left( {1.20 \times {{10}^6}\,{\rm{V/m}}} \right)\\\\ = 3.82 \times {10^{ - 5}}\,\,{{\rm{C}}^{\rm{2}}}{\rm{/N}} \cdot {{\rm{m}}^{\rm{2}}}\\\end{array}$

(b)

Calculate the magnitude of charge per unit area on the surface of dielectric plate.

The expression for magnitude of charge per unit area on the surface of dielectric plate is,

$\sigma ' = \sigma \left( {1 - \frac{1}{k}} \right)$

Here, $\sigma$ is the charge density of conducting plate ${\sigma _1}$ is the charge density on surface of dielectric plate, and k is the dielectric constant.

Substitute $3.82 \times {10^{ - 5}}\,{{\rm{C}}^{\rm{2}}}{\rm{/N}} \cdot {{\rm{m}}^{\rm{2}}}$ for $\sigma$ and $3.60$ for k.

$\begin{array}{c}\\\sigma ' = 3.82 \times {10^{ - 5}}\,{{\rm{C}}^{\rm{2}}}{\rm{/N}} \cdot {{\rm{m}}^{\rm{2}}}\left( {1 - \frac{1}{{3.60}}} \right)\\\\ = 2.76 \times {10^{ - 5}}\,{\rm{C/}}{{\rm{m}}^{\rm{2}}}\\\end{array}$

(c)

Calculate the total electric- field energy stored in the capacitor by using the expression as follows:

$U = \frac{1}{2}k{\varepsilon _0}{E^2}Ad$

Convert the area from centimeter square to meter square by multiplying it with${10^{ - 4}}$.

$\begin{array}{c}\\A = \left( {2.5\,\,{\rm{c}}{{\rm{m}}^{\rm{2}}}} \right)\left( {\frac{{{{10}^{ - 4}}\,{{\rm{m}}^{\rm{2}}}}}{{1\,\,{\rm{c}}{{\rm{m}}^{\rm{2}}}}}} \right)\\\\ = 2.5 \times {10^{ - 4}}\,\,{{\rm{m}}^{\rm{2}}}\\\end{array}$

Convert the diameter from millimeter to meter by multiplying it with ${10^{ - 3}}$.

$\begin{array}{c}\\d = \left( {1.8\,{\rm{mm}}} \right)\left( {\frac{{{{10}^{ - 3}}\,{\rm{m}}}}{{1\,\,{\rm{mm}}}}} \right)\\\\ = 1.8 \times {10^{ - 3}}\,{\rm{m}}\\\end{array}$

Substitute 3.60 for k , $8.85 \times {10^{ - 12}}\,{{\rm{C}}^{\rm{2}}}{\rm{/N}} \cdot {{\rm{m}}^{\rm{2}}}$ for ${\varepsilon _0}$,$1.20 \times {10^6}\,{\rm{V/m}}$for $E$, $2.5 \times {10^{ - 4}}\,{{\rm{m}}^{\rm{2}}}$ for $A$, and $1.8 \times {10^{ - 3}}\,{\rm{m}}$ for $d$ in expression $U = \frac{1}{2}k{\varepsilon _0}{E^2}Ad$.

$\begin{array}{c}\\U = \frac{1}{2}\left( {3.60} \right)\left( {8.85 \times {{10}^{ - 12}}\,{{\rm{C}}^{\rm{2}}}{\rm{/N}} \cdot {{\rm{m}}^{\rm{2}}}} \right){\left( {1.20 \times {{10}^6}\,{\rm{V/m}}} \right)^2}\left( {2.5 \times {{10}^{ - 4}}\,{{\rm{m}}^{\rm{2}}}} \right)\left( {1.8 \times {{10}^{ - 3}}\,{\rm{m}}} \right)\\\\ = 1.03 \times {10^{ - 5}}\,{\rm{J}}\\\end{array}$

Ans: Part a

The magnitude of charge per unit area on the conducting plate is $3.82 \times {10^{ - 5}}\,{{\rm{C}}^{\rm{2}}}{\rm{/N}} \cdot {{\rm{m}}^{\rm{2}}}$.

Part b

The magnitude of charge per unit area on the surface of dielectric plate is $2.76 \times {10^{ - 5}}\,{\rm{C/}}{{\rm{m}}^{\rm{2}}}$.

Part c

The total electric field-energy stored in the capacitor is $1.03 \times {10^{ - 5}}\,{\rm{J}}$.