Energy Stored In Capacitors, Field In Dielectrics, Gauss’S Law In Dielectrics - Electrostatic Energy Store In A Capacitor

Slide 1: Energy Stored in Capacitors

Capacitors are used to store electrical energy.
The energy stored in a capacitor can be calculated using the formula:
- Energy (E) = 0.5 * C * V^2, where C is the capacitance and V is the voltage across the capacitor.
The unit of energy is Joules (J).

Slide 2: Capacitance

Capacitance (C) is a measure of a capacitor’s ability to store charge.
It is defined as the ratio of the charge stored in the capacitor (Q) to the voltage across it (V).
- C = Q / V
The unit of capacitance is Farad (F).
Capacitance depends on the physical characteristics of the capacitor, such as its geometry and the material used.

Slide 3: Field in Dielectrics

When a dielectric material is placed between the plates of a capacitor, it affects the electric field.
Dielectric materials are insulators that can be polarized by an external electric field.
The presence of a dielectric material reduces the strength of the electric field between the plates.
The electric field in a dielectric material is given by:
- E = E0 / k, where E0 is the electric field without the dielectric and k is the dielectric constant.

Slide 4: Dielectric Constant

The dielectric constant (k) is a measure of a dielectric material’s ability to store electrical energy.
It is defined as the ratio of the electric field without the dielectric (E0) to the electric field in the dielectric (E).
The dielectric constant varies for different materials and is dimensionless.
The greater the dielectric constant, the higher the capacitance of the capacitor.

Slide 5: Gauss’s Law in Dielectrics

Gauss’s law states that the electric flux through a closed surface is equal to the total charge enclosed divided by the permittivity of the material.
In the presence of dielectric material, the relationship becomes:
- ∮ E * dA = (Q’ - Q) / ε, where E is the electric field, A is the area, Q’ is the free charge, Q is the bound charge, and ε is the permittivity of the dielectric material.
The presence of the dielectric material affects the enclosed charge and thus the electric flux.

Slide 6: Electrostatic Energy Store in a Capacitor

The energy stored in a capacitor with a dielectric can be calculated using:
- Energy (E) = 0.5 * C * V^2 for a capacitor without a dielectric.
- Energy (E) = 0.5 * C * V^2 / k for a capacitor with a dielectric, where k is the dielectric constant.
The presence of a dielectric material increases the energy stored in the capacitor compared to without the dielectric.
The energy stored can be used to do work or power electronic devices.

Slide 7: Example - Energy Stored in a Capacitor

Let’s consider a capacitor with a capacitance of 10 μF (microfarads) and a voltage of 12 V.
Using the formula for energy stored in a capacitor, we can calculate:
- Energy (E) = 0.5 * 10e-6 F * (12 V)^2
- Energy = 0.72 mJ (millijoules)
This means that the capacitor can store 0.72 millijoules of energy.

Slide 8: Example - Capacitance Calculation

Suppose we have a parallel plate capacitor with a plate area of 100 cm² and a plate separation distance of 0.5 mm.
The capacitance can be calculated using the formula:
- C = ε₀ * A / d, where ε₀ is the permittivity of free space, A is the plate area, and d is the plate separation distance.
Substituting the given values, we have:
- C = (8.85e-12 F/m) * (10e-4 m²) / (0.5e-3 m)
- C = 0.177 F (Farads)
Therefore, the capacitance of the parallel plate capacitor is 0.177 Farads.

Slide 9: Example - Field in Dielectric

Consider a capacitor with a voltage of 20 V and a dielectric material with a dielectric constant of 4.
The electric field in the dielectric can be calculated using the formula:
- E = E₀ / k
- E = 20 V / 4
- E = 5 V/m (Volts per meter)
Therefore, the electric field in the dielectric material is 5 Volts per meter.

Slide 10: Example - Gauss’s Law in Dielectrics

Suppose we have a closed surface enclosing a charge of 5 μC (microcoulombs) in the presence of a dielectric with a permittivity of 8.85e-12 C²/N·m².
Using Gauss’s law in dielectrics, we can calculate the electric flux through the surface:
- ∮ E * dA = (Q’ - Q) / ε
- ∮ E * dA = (0 - 5e-6 C) / (8.85e-12 C²/N·m²)
- ∮ E * dA = -5.65e5 N·m²/C
The electric flux through the surface is -5.65e5 N·m²/C.

Slide 11: Energy Stored in Capacitors (contd.)

The energy stored in a capacitor can also be calculated using the formula:
- Energy (E) = 0.5 * Q * V, where Q is the charge stored in the capacitor and V is the voltage across it.
This formula is derived from the capacitance formula and Ohm’s Law.
The energy stored in a capacitor can be released quickly, making it useful in applications such as camera flashes and defibrillators.

Slide 12: Example - Energy Stored in a Capacitor (contd.)

Let’s consider a capacitor with a charge of 20 μC (microcoulombs) and a voltage of 5 V.
Using the formula for energy stored in a capacitor, we can calculate:
- Energy (E) = 0.5 * (20e-6 C) * (5 V)
- Energy = 0.5 mJ (millijoules)
This means that the capacitor can store 0.5 millijoules of energy.

Slide 13: Field in Dielectrics (contd.)

The electric field in a dielectric material is weaker than the electric field in free space or vacuum.
The presence of the dielectric material increases the capacitance of the capacitor.
Dielectric materials are commonly used in capacitors to increase their storage capacity.
The dielectric material also provides insulation, preventing the flow of current between the capacitor plates.

Slide 14: Dielectric Constant (contd.)

The dielectric constant of a material is a measure of how well it can store electrical energy compared to free space.
Materials with high dielectric constants are used in capacitors to increase their capacitance.
Some common materials with high dielectric constants include ceramic, tantalum, and aluminum electrolytic.
Dielectric constants can vary significantly for different materials, allowing for a wide range of capacitance values in capacitors.

Slide 15: Gauss’s Law in Dielectrics (contd.)

Gauss’s law in dielectrics relates the electric field, charge, and permittivity of the dielectric material.
The equation is given by:
- ∮ E * dA = (Q’ - Q) / ε, where E is the electric field, A is the area, Q’ is the free charge, Q is the bound charge, and ε is the permittivity of the dielectric material.
Bound charges are charges within the dielectric material that are influenced by the external electric field.
The presence of bound charges affects the electric flux through a closed surface, as described by Gauss’s law.

Slide 16: Example - Gauss’s Law in Dielectrics (contd.)

Suppose we have a closed surface enclosing a charge of 10 μC (microcoulombs) in the presence of a dielectric with a permittivity of 5e-11 C²/N·m².
Using Gauss’s law in dielectrics, we can calculate the electric flux through the surface:
- ∮ E * dA = (Q’ - Q) / ε
- ∮ E * dA = (0 - 10e-6 C) / (5e-11 C²/N·m²)
- ∮ E * dA = -2e5 N·m²/C
The electric flux through the surface is -2e5 N·m²/C.

Slide 17: Electrostatic Energy Store in a Capacitor (contd.)

The energy stored in a capacitor can also be understood as the work done to charge the capacitor.
When a capacitor is charged, work is done to move charges against the electric field.
This work is stored as potential energy in the electric field between the capacitor plates.
The energy stored can also be calculated using the formula:
- Energy (E) = Q * V / 2, where Q is the charge stored in the capacitor and V is the voltage across it.

Slide 18: Example - Electrostatic Energy Store in a Capacitor (contd.)

Let’s consider a capacitor with a charge of 15 μC (microcoulombs) and a voltage of 8 V.
Using the formula for energy stored in a capacitor, we can calculate:
- Energy (E) = (15e-6 C) * (8 V) / 2
- Energy = 0.06 J (Joules)
This means that the capacitor can store 0.06 Joules of energy.

Slide 19: Introduction to Magnetic Fields

Magnetic fields are created by moving charged particles.
They are different from electric fields, which are created by stationary charges.
Magnetic fields can exert forces on moving charges and other magnetic objects.
The strength and direction of a magnetic field are represented by magnetic field lines.
Magnetic field lines form closed loops and are directed from north to south.

Slide 20: Magnetic Field Strength (Magnetic Field Intensity)

The strength of a magnetic field at a point is measured by its magnetic field strength.
The magnetic field strength is given by the formula:
- B = μ₀ * (I / 2πr), where B is the magnetic field strength, μ₀ is the permeability of free space, I is the current, and r is the distance from the wire.
The unit of magnetic field strength is Tesla (T).
The magnetic field strength is inversely proportional to the distance from the wire.

Slide 21: Energy Stored in Capacitors (contd.)

The energy stored in a capacitor is proportional to the square of the voltage across it.
Doubling the voltage across a capacitor will quadruple the energy stored.
The energy stored in a capacitor is always positive, regardless of the charge polarity or the direction of the voltage.

Slide 22: Field in Dielectrics (contd.)

The presence of a dielectric material increases the capacitance of a capacitor without changing the charge, voltage, or energy stored.
Dielectric materials can be polarized by an applied electric field, aligning their positive and negative charges in opposite directions.
The polarization of the dielectric causes an additional field, which reduces the effective field between the capacitor plates.

Slide 23: Gauss’s Law in Dielectrics (contd.)

Gauss’s law in dielectrics applies to both closed and open surfaces.
For an open surface, the equation becomes:
- E * A = Q’ / ε, where E is the electric field, A is the area, Q’ is the free charge, and ε is the permittivity of the dielectric.
This equation describes the relationship between the electric field and the charge outside the surface of a dielectric.

Slide 24: Electrostatic Energy Store in a Capacitor (contd.)

The energy stored in a capacitor is directly proportional to the square of the charge stored.
Doubling the charge stored in a capacitor will quadruple the energy stored.
The energy stored in a capacitor can also be calculated using the equation:
- Energy (E) = Q^2 / (2C), where Q is the charge stored in the capacitor and C is the capacitance.

Slide 25: Example - Energy Stored in a Capacitor (contd.)

Let’s consider a capacitor with a charge of 30 μC (microcoulombs) and a capacitance of 4 μF (microfarads).
Using the formula for energy stored in a capacitor, we can calculate:
- Energy (E) = (30e-6 C)^2 / (2 * 4e-6 F)
- Energy = 0.675 J (Joules)
This means that the capacitor can store 0.675 Joules of energy.

Slide 26: Example - Capacitance Calculation (contd.)

Suppose we have a cylindrical capacitor with a radius of 5 cm and a length of 10 cm.
The capacitance can be calculated using the formula:
- C = (2πε₀ / ln(b/a)) * L, where ε₀ is the permittivity of free space, a is the inner radius, b is the outer radius, and L is the length of the capacitor.
Substituting the given values, we have:
- C = (2π * 8.85e-12 C²/N·m² / ln(0.05/0.025)) * 0.1 m
- C = 427.67 pF (picofarads)
Therefore, the capacitance of the cylindrical capacitor is 427.67 picofarads.

Slide 27: Example - Field in Dielectric (contd.)

Consider a parallel plate capacitor with a voltage of 12 V and a dielectric material with a dielectric constant of 3.
The electric field in the dielectric can be calculated using the formula:
- E = E₀ / k
- E = 12 V / 3
- E = 4 V/m (Volts per meter)
Therefore, the electric field in the dielectric material is 4 Volts per meter.

Slide 28: Example - Gauss’s Law in Dielectrics (contd.)

Suppose we have an open surface with an area of 10 cm² and a free charge of 8 μC (microcoulombs) in the presence of a dielectric with a permittivity of 7e-11 C²/N·m².
Using Gauss’s law in dielectrics, we can calculate the electric field:
- E * A = Q’ / ε
- (E) * (10e-4 m²) = 8e-6 C / (7e-11 C²/N·m²)
- E = 1.14e6 N/C (Newtons per Coulomb)
The electric field is 1.14e6 N/C.

Slide 29: Example - Electrostatic Energy Store in a Capacitor (contd.)

Let’s consider a capacitor with a charge of 25 μC (microcoulombs) and a capacitance of 8 μF (microfarads).
Using the formula for energy stored in a capacitor, we can calculate:
- Energy (E) = (25e-6 C)^2 / (2 * 8e-6 F)
- Energy = 0.78125 J (Joules)
This means that the capacitor can store 0.78125 Joules of energy.

Slide 30: Recap and Summary

Energy is stored in capacitors and can be calculated using various formulas.
Capacitance is a measure of a capacitor’s ability to store charge and is influenced by the physical characteristics of the capacitor.
The presence of a dielectric material increases the capacitance and reduces the strength of the electric field in the capacitor.
Gauss’s law in dielectrics relates the electric field, charge, and permittivity of the dielectric material.
Understanding the energy stored in capacitors and the behavior of fields in dielectrics is essential for various applications and circuits.