Energy Density in Electrostatic Field MCQ Quiz - Objective Question with Answer for Energy Density in Electrostatic Field - Download Free PDF

Concept

Series connection of capacitors:

When 'n' capacitors are connected in series, the equivalent capacitance is given by:

\({1\over C_{eq}}={1\over C_{1}}+{1\over C_{2}}........{1\over C_{n}}\)

If all capacitors are equal, then:

\(C_{eq}={C\over n}\)

Parallel connection of capacitors:

When 'n' capacitors are connected in parallel, the equivalent capacitance is given by:

\(C_{eq}=C_1+C_2.......C_n\)

If all capacitors are equal, then:

\(C_{eq}=nC\)

Calculation

Given, n = 10

C_eq = 10 μF

\(10={C\over 10}\)

\(C=100\space \mu F\)

When capacitors are connected in parallel.

\(C_{eq}=10\times 100\times 10^{-6}\)

C_eq = 10^-3 F

The energy in the capacitor is given by:

\(E={1\over 2}CV^2\)

\(E={1\over 2}\times 10^{-3}\times (100)^2\)

E = 5 J

Co-energy in a magnetic circuit:

• Co-energy is the dual of energy, a non-physical quantity useful for theoretical analysis of systems storing and transforming energy.
• Co-energy is expressed in the same units as energy and is especially useful for the calculation of magnetic forces and torque in rotating machines.
• The value of co-energy is zero for the system which is incapable of storing energy.

Co-energy in a linear magnetic circuit:

For a single excited linear magnetic circuit, the co-energy is equal to the field energy.

From the above graph, OBAO (field energy) = OACA (co-energy)

Calculation:

The co-energy for a parallel plate capacitor is given by:

\(W = { 1\over 2}CV^2\)

where, C = Capacitance

V = Voltage 

W = Co-energy

∵ \(C = {\epsilon_oA \over d}\)

\(W = { 1\over 2}( {\epsilon_oA \over d})V^2\)

\(W = { 1\over 2}( {\epsilon_oA \over d})(Ed)^2\)

Co-energy density is the ratio of energy to the energy to volume (A × d)

W = \( {1 \over 2}\)E2ε0

Concept:

Energy stored in capacitor:

• A capacitor is a device to store energy.
• The process of charging up a capacitor involves the transferring of electric charges from one plate to another.
• The work done in charging the capacitor is stored as its electrical potential energy.
• The energy stored in the capacitor is
   

\(U = \frac{1}{2}\frac{{{Q^2}}}{C} = \frac{1}{2}C{V^2} = \frac{1}{2}QV\)

Where Q = charge stored on the capacitor,

U = energy stored in the capacitor,

C = capacitance of the capacitor and

V = Electric potential difference

When capacitors are connected in parallel, the total capacitance is the sum of the individual capacitors' capacitances.

\({C_{eq}}(parallel) = {C_1} + {C_2} + {C_3} + \ldots + {C_4}\)

When capacitors are connected in series, the total capacitance is less than the least capacitance connected in series.

\(\frac{1}{{{C_{eq}(series)}}} = \frac{1}{{{C_1}}} + \frac{1}{{{C_2}}} + \ldots + \frac{1}{{{C_n}}}\)

Calculations:

C_eq = 4 μF, V = 400

When capacitances are connected in series

then, \(\frac{1}{{{C_{eq}}}} = \frac{5}{C}\)

C = 20 μF

When capacitances are connected in parallel

C_eq' = 20 × 5 = 100 μF

\(U = \frac{{100 \times {{10}^{ - 6}} \times 400 \times 400}}{2}\)

= 8 J

Concept:

Energy stored by the inductor (E_L) is given by,

\(E_L=\frac{1}{2}L{I^2}\)

Energy stored by a capacitor (E_C) is given by,

\(E_C =\frac{1}{2}C{V^2}\)

Calculation: 

Given: L = 2 H , I = 4 A,  V_C  = 400 V 

From the above concept,

\(\frac{1}{2}L{I^2} = \frac{1}{2}C{V^2}\)

\(L{I^2} = C{V^2}\)

\(2 \times {\left( 4 \right)^2} = C{\left( {400} \right)^2}\)

\(C = \frac{{32}}{{160000}}\)

\(C = 200\mu F\)

Concept:

Electrical energy density = total energy per unit volume

\({W_e} = \frac{1}{2}\;\vec Q \cdot \vec E = \frac{1}{2}\varepsilon {E^2}\;J/{m^3}\)

Where ε → Medium permittivity

E → Electric field or potential gradient.

Calculation:

Given that the medium is air.

ε = 8.854 × 10^-12 F/m

Potential gradient, E = 30 × 10³ × 10² V/m

= 3 × 10⁶ V/m

\({W_e} = \frac{1}{2} \times 8.854 \times {10^{ - 12}} \times 3 \times {10^6} \times 3 \times {10^6} = 40\;J/{m^3}\)

Concept:

Energy stored by the inductor (E_L) is given by,

\(E_L=\frac{1}{2}L{I^2}\)

Energy stored by a capacitor (E_C) is given by,

\(E_C =\frac{1}{2}C{V^2}\)

Calculation: 

Given: L = 2 H , I = 4 A,  V_C  = 400 V 

From the above concept,

\(\frac{1}{2}L{I^2} = \frac{1}{2}C{V^2}\)

\(L{I^2} = C{V^2}\)

\(2 \times {\left( 4 \right)^2} = C{\left( {400} \right)^2}\)

\(C = \frac{{32}}{{160000}}\)

\(C = 200\mu F\)

Concept:

Energy stored in capacitor:

• A capacitor is a device to store energy.
• The process of charging up a capacitor involves the transferring of electric charges from one plate to another.
• The work done in charging the capacitor is stored as its electrical potential energy.
• The energy stored in the capacitor is
   

\(U = \frac{1}{2}\frac{{{Q^2}}}{C} = \frac{1}{2}C{V^2} = \frac{1}{2}QV\)

Where Q = charge stored on the capacitor,

U = energy stored in the capacitor,

C = capacitance of the capacitor and

V = Electric potential difference

When capacitors are connected in parallel, the total capacitance is the sum of the individual capacitors' capacitances.

\({C_{eq}}(parallel) = {C_1} + {C_2} + {C_3} + \ldots + {C_4}\)

When capacitors are connected in series, the total capacitance is less than the least capacitance connected in series.

\(\frac{1}{{{C_{eq}(series)}}} = \frac{1}{{{C_1}}} + \frac{1}{{{C_2}}} + \ldots + \frac{1}{{{C_n}}}\)

Calculations:

C_eq = 4 μF, V = 400

When capacitances are connected in series

then, \(\frac{1}{{{C_{eq}}}} = \frac{5}{C}\)

C = 20 μF

When capacitances are connected in parallel

C_eq' = 20 × 5 = 100 μF

\(U = \frac{{100 \times {{10}^{ - 6}} \times 400 \times 400}}{2}\)

= 8 J

Concept:

In a parallel plate capacitor, the energy density is given by

\(\vec E = \frac{1}{2}\frac{{{{\left| D \right|}^2}}}{\varepsilon }\)

D is the flux density

ε is permittivity

Calculation:

Flux density \(\vec D = 15\;\mu C/{m^2}\)

= 15 × 10^-6 C/m²

Energy density \(\vec E = 20\;J/{m^3}\)

To find relative permittivity ε_r:

We know that,

\(\Rightarrow 20\frac{J}{{{m^3}}} = \frac{1}{2} \times \frac{{{{\left( {15\; \times \;{{10}^{ - 6}}} \right)}^2}}}{{\frac{1}{{36\pi }}\; \times \;{{10}^{ - 9}}{\varepsilon _r}}}\)

\(\Rightarrow {\varepsilon _r} = \frac{{15\; \times \;15\; \times \;{{10}^{ - 12}}\; \times \;36\pi }}{{2\; \times \;20\; \times\; {{10}^{ - 9}}}}\)

⇒ ε_r = 0.635 ≈ 0.6

Co-energy in a magnetic circuit:

• Co-energy is the dual of energy, a non-physical quantity useful for theoretical analysis of systems storing and transforming energy.
• Co-energy is expressed in the same units as energy and is especially useful for the calculation of magnetic forces and torque in rotating machines.
• The value of co-energy is zero for the system which is incapable of storing energy.

Co-energy in a linear magnetic circuit:

For a single excited linear magnetic circuit, the co-energy is equal to the field energy.

From the above graph, OBAO (field energy) = OACA (co-energy)

Calculation:

The co-energy for a parallel plate capacitor is given by:

\(W = { 1\over 2}CV^2\)

where, C = Capacitance

V = Voltage 

W = Co-energy

∵ \(C = {\epsilon_oA \over d}\)

\(W = { 1\over 2}( {\epsilon_oA \over d})V^2\)

\(W = { 1\over 2}( {\epsilon_oA \over d})(Ed)^2\)

Co-energy density is the ratio of energy to the energy to volume (A × d)

W = \( {1 \over 2}\)E2ε0

Concept:  

Capacitance with the separation is r is given by \({{\rm{C}}} = \frac{{{\rm{\epsilon A}}}}{{{\rm{r}}}}\)

Application:

Capacitance when the separation is 2d.

\({{\rm{C}}_1} = \frac{{{\rm{\epsilon A}}}}{{2{\rm{d}}}}\)

Let the charge on plates be Q coulomb. Then, the energy of the capacitor with separation 2d is

\(\begin{array}{l} {{\rm{E}}_1} = \frac{{{{\rm{Q}}^2}}}{{2{{\rm{C}}_1}}}\\ {{\rm{E}}_1} = \frac{{{{\rm{Q}}^2}}}{{2{{\rm{C}}_1}}} = \frac{{{{\rm{Q}}^2}2{\rm{d}}}}{{2{\rm{\epsilon A}}}} = \frac{{{{\rm{Q}}^2}{\rm{d}}}}{{{\rm{\epsilon A}}}} \end{array}\)

The energy of the capacitor with separation d is

Now,   \({\rm{C}} = \frac{{{\rm{A}}}}{{\rm{d}}} \Rightarrow {\rm{E}} = \frac{{{{\rm{Q}}^2}{\rm{d}}}}{{2{\rm{\epsilon A}}}}\)

\(\Rightarrow {\rm{E}} = \frac{{{{\rm{Q}}^2}{\rm{d}}}}{{2{\rm{\epsilon A}}}}\)

Thus, we see \({\rm{E}} = \frac{1}{2}{{\rm{E}}_1}\)

Concept

Series connection of capacitors:

When 'n' capacitors are connected in series, the equivalent capacitance is given by:

\({1\over C_{eq}}={1\over C_{1}}+{1\over C_{2}}........{1\over C_{n}}\)

If all capacitors are equal, then:

\(C_{eq}={C\over n}\)

Parallel connection of capacitors:

When 'n' capacitors are connected in parallel, the equivalent capacitance is given by:

\(C_{eq}=C_1+C_2.......C_n\)

If all capacitors are equal, then:

\(C_{eq}=nC\)

Calculation

Given, n = 10

C_eq = 10 μF

\(10={C\over 10}\)

\(C=100\space \mu F\)

When capacitors are connected in parallel.

\(C_{eq}=10\times 100\times 10^{-6}\)

C_eq = 10^-3 F

The energy in the capacitor is given by:

\(E={1\over 2}CV^2\)

\(E={1\over 2}\times 10^{-3}\times (100)^2\)

E = 5 J

Concept:

Energy stored by the inductor (E_L) is given by,

\(E_L=\frac{1}{2}L{I^2}\)

Energy stored by a capacitor (E_C) is given by,

\(E_C =\frac{1}{2}C{V^2}\)

Calculation: 

Given: L = 2 H , I = 4 A,  V_C  = 400 V 

From the above concept,

\(\frac{1}{2}L{I^2} = \frac{1}{2}C{V^2}\)

\(L{I^2} = C{V^2}\)

\(2 \times {\left( 4 \right)^2} = C{\left( {400} \right)^2}\)

\(C = \frac{{32}}{{160000}}\)

\(C = 200\mu F\)

Concept:

Energy stored in capacitor:

• A capacitor is a device to store energy.
• The process of charging up a capacitor involves the transferring of electric charges from one plate to another.
• The work done in charging the capacitor is stored as its electrical potential energy.
• The energy stored in the capacitor is
   

\(U = \frac{1}{2}\frac{{{Q^2}}}{C} = \frac{1}{2}C{V^2} = \frac{1}{2}QV\)

Where Q = charge stored on the capacitor,

U = energy stored in the capacitor,

C = capacitance of the capacitor and

V = Electric potential difference

When capacitors are connected in parallel, the total capacitance is the sum of the individual capacitors' capacitances.

\({C_{eq}}(parallel) = {C_1} + {C_2} + {C_3} + \ldots + {C_4}\)

When capacitors are connected in series, the total capacitance is less than the least capacitance connected in series.

\(\frac{1}{{{C_{eq}(series)}}} = \frac{1}{{{C_1}}} + \frac{1}{{{C_2}}} + \ldots + \frac{1}{{{C_n}}}\)

Calculations:

C_eq = 4 μF, V = 400

When capacitances are connected in series

then, \(\frac{1}{{{C_{eq}}}} = \frac{5}{C}\)

C = 20 μF

When capacitances are connected in parallel

C_eq' = 20 × 5 = 100 μF

\(U = \frac{{100 \times {{10}^{ - 6}} \times 400 \times 400}}{2}\)

= 8 J

Concept:

Electrical energy density = total energy per unit volume

\({W_e} = \frac{1}{2}\;\vec Q \cdot \vec E = \frac{1}{2}\varepsilon {E^2}\;J/{m^3}\)

Where ε → Medium permittivity

E → Electric field or potential gradient.

Calculation:

Given that the medium is air.

ε = 8.854 × 10^-12 F/m

Potential gradient, E = 30 × 10³ × 10² V/m

= 3 × 10⁶ V/m

\({W_e} = \frac{1}{2} \times 8.854 \times {10^{ - 12}} \times 3 \times {10^6} \times 3 \times {10^6} = 40\;J/{m^3}\)

Concept:

In a parallel plate capacitor, the energy density is given by

\(\vec E = \frac{1}{2}\frac{{{{\left| D \right|}^2}}}{\varepsilon }\)

D is the flux density

ε is permittivity

Calculation:

Flux density \(\vec D = 15\;\mu C/{m^2}\)

= 15 × 10^-6 C/m²

Energy density \(\vec E = 20\;J/{m^3}\)

To find relative permittivity ε_r:

We know that,

\(\Rightarrow 20\frac{J}{{{m^3}}} = \frac{1}{2} \times \frac{{{{\left( {15\; \times \;{{10}^{ - 6}}} \right)}^2}}}{{\frac{1}{{36\pi }}\; \times \;{{10}^{ - 9}}{\varepsilon _r}}}\)

\(\Rightarrow {\varepsilon _r} = \frac{{15\; \times \;15\; \times \;{{10}^{ - 12}}\; \times \;36\pi }}{{2\; \times \;20\; \times\; {{10}^{ - 9}}}}\)

⇒ ε_r = 0.635 ≈ 0.6