Potential Due To Different Charge Distributions

Definition of Electron Volt (eV)

- In the previous lectures, we discussed the concept of potential difference and electric potential. - The potential difference between two points in an electric field is defined as the amount of work done per unit charge in moving a test charge from one point to another. - In this lecture, we will extend our understanding of potential to different charge distributions. - We will also introduce the concept of the electron volt (eV) as a unit of energy. - The electric potential at a point due to a point charge is given by Coulomb's law. - For a point charge $ Q $ at a distance $ r $ from the point, the electric potential $ V $ is given by the equation: $ V = \frac{kQ}{r} $ Where $ k $ is the electrostatic constant. - The potential due to a point charge decreases as the distance from the charge increases. - Now, let's consider the potential due to a collection of point charges. - The total potential at a point due to multiple charges is the algebraic sum of the potentials due to each individual charge. - If there are $ n $ charges $ Q_i $ at distances $ r_i $ from the point, the total potential $ V $ at that point is given by: $ V = \frac{kQ_1}{r_1} + \frac{kQ_2}{r_2} + \frac{kQ_3}{r_3} + \ldots + \frac{kQ_n}{r_n} $ - The principle of superposition holds true for electric potential. - When dealing with continuous charge distributions, we need to use integrals to find the potential at a point. - For a linear charge density $ \lambda(x) $ , the potential $ dV $ at a point $ P $ due to a small segment $ dx $ is given by: $ dV = \frac{k\lambda(x)}{r}dx $ - To find the potential due to the entire distribution, we integrate the above equation over the entire length of the distribution. - - For a surface charge density $ \sigma(x, y) $ , the potential $ dV $ at a point $ P $ due to an element $ dA $ is given by: $ dV = \frac{k\sigma(x, y)}{r}dA $ - To find the potential due to the entire surface, we integrate the above equation over the entire surface area.

For a volume charge density $ \rho(x, y, z) $ , the potential $ dV $ at a point $ P $ due to a small volume element $ d\tau $ is given by: $ dV = \frac{k\rho(x, y, z)}{r}d\tau $
To find the potential due to the entire volume, we integrate the above equation over the entire volume.
We can also define the electric potential using the electric field ( $ E $ ) at a point.
The electric field at a point due to a charge distribution is defined as the force experienced by a unit positive test charge placed at that point.
The potential at a point is the work done in bringing a unit positive test charge from infinity to that point against the electric field.
We can use the electric field equation to find the potential due to a charge distribution.
Now, let’s introduce the concept of the electron volt (eV).
The electron volt is a unit of energy commonly used in atomic and nuclear physics.
It is defined as the amount of energy gained or lost by an electron when it is accelerated through an electric potential difference of one volt.
1 eV is equivalent to the charge of an electron ( $ 1.6 \times 10^{-19} $ C) multiplied by one volt.
The electron volt is a convenient unit for measuring energies on an atomic scale.
Energy is quantized in atomic and subatomic systems, and the electron volt allows us to easily express these energy levels.
For example, the ionization energy of hydrogen is 13.6 eV.
It represents the minimum amount of energy required to ionize hydrogen atom, i.e., to remove an electron from the atom.
In summary, in this lecture, we discussed the potential due to different charge distributions.
We explained the concept of the electron volt as a unit of energy.
We also learned how to calculate the potential due to point charges, linear charge distributions, surface charge distributions, and volume charge distributions.
Next, we will solve numerical problems to reinforce our understanding of these concepts.
Stay tuned for the next lecture!

Potential Due to Point Charges:

The electric potential at a point due to a point charge is given by Coulomb’s law:
- $ V = \frac{kQ}{r} $
The potential decreases as the distance from the charge increases.

Potential Due to Multiple Charges:

The total potential at a point due to multiple charges is the sum of the potentials due to each individual charge.
Use the principle of superposition:
- $ V = \frac{kQ_1}{r_1} + \frac{kQ_2}{r_2} + \frac{kQ_3}{r_3} + \ldots + \frac{kQ_n}{r_n} $

Potential Due to Linear Charge Distribution:

For a linear charge density $ \lambda(x) $ , the potential at a point $ P $ due to a small segment $ dx $ is given by:
- $ dV = \frac{k\lambda(x)}{r}dx $
To find the potential due to the entire distribution, integrate the above equation over the entire length.

Potential Due to Surface Charge Distribution:

For a surface charge density $ \sigma(x, y) $ , the potential at a point $ P $ due to an element $ dA $ is given by:
- $ dV = \frac{k\sigma(x, y)}{r}dA $
To find the potential due to the entire surface, integrate the above equation over the entire surface area.

Potential Due to Volume Charge Distribution:

For a volume charge density $ \rho(x, y, z) $ , the potential at a point $ P $ due to a small volume element $ d\tau $ is given by:
- $ dV = \frac{k\rho(x, y, z)}{r}d\tau $
To find the potential due to the entire volume, integrate the above equation over the entire volume.

Potential due to Electric Field:

The electric field ( $ E $ ) at a point is defined as the force experienced by a unit positive test charge placed at that point.
The potential at a point is the work done in bringing a unit positive test charge from infinity to that point against the electric field.

Electron Volt (eV):

The electron volt is a unit of energy commonly used in atomic and nuclear physics.
It is defined as the amount of energy gained or lost by an electron when it is accelerated through an electric potential difference of one volt.
1 eV is equal to the charge of an electron ( $ 1.6 \times 10^{-19} $ C) multiplied by one volt.

Applications of Electron Volt:

The electron volt is used to measure energy levels in atomic and subatomic systems.
It is particularly useful for quantizing energy levels and understanding processes such as ionization and excitation.
For example, the ionization energy of hydrogen is 13.6 eV, which represents the energy required to remove an electron from a hydrogen atom.

Calculating Energy using eV:

To calculate energy using electron volts, multiply the charge in electron volts by the potential difference in volts.
For example, if the charge is 2 eV and the potential difference is 5 V, the energy would be:
- $ E = 2 \times 5 $ eV

Summary:

In this lecture, we discussed the potential due to different charge distributions, including point charges, linear charge distributions, surface charge distributions, and volume charge distributions.
We also learned about the electron volt as a unit of energy.
Next, we will solve numerical problems to solidify our understanding of these concepts.

Slide 21

Potential Due to Different Charge Distributions: Summary
Potential due to point charges:
- Given by Coulomb’s law: $ V = \frac{kQ}{r} $
Potential due to multiple charges:
- Principle of superposition: $ V = \frac{kQ_1}{r_1} + \frac{kQ_2}{r_2} + \frac{kQ_3}{r_3} + \ldots + \frac{kQ_n}{r_n} $
Potential due to linear charge distribution:
- For linear charge density $ \lambda(x) $ : $ dV = \frac{k\lambda(x)}{r}dx $
Potential due to surface charge distribution:
- For surface charge density $ \sigma(x, y) $ : $ dV = \frac{k\sigma(x, y)}{r}dA $

Slide 22

Potential due to volume charge distribution:
- For volume charge density $ \rho(x, y, z) $ : $ dV = \frac{k\rho(x, y, z)}{r}d\tau $
Potential due to electric field:
- Electric field ( $ E $ ) at a point is the force experienced by a unit positive test charge
- Potential at a point is the work done to bring a unit positive test charge from infinity against the electric field
Electron Volt (eV):
- Unit of energy equal to the charge of an electron ( $ 1.6 \times 10^{-19} $ C) multiplied by one volt

Slide 23

Electron Volt (eV):
- Used for measuring energy on an atomic and subatomic scale
- Allows easy expression of energy levels and processes such as ionization
Example: Ionization Energy of Hydrogen:
- Ionization energy = 13.6 eV
- Represents the energy required to remove an electron from a hydrogen atom
Calculating Energy using eV:
- Energy ( $ E $ ) = charge ( $ Q $ ) $ \times $ potential difference ( $ V $ )
- For example, if charge = 2 eV and potential difference = 5 V, then energy ( $ E $ ) = 2 $ \times $ 5 eV

Slide 24

Potential Due to Different Charge Distributions: Summary
Potential due to point charges:
- $ V = \frac{kQ}{r} $
Potential due to multiple charges:
- $ V = \frac{kQ_1}{r_1} + \frac{kQ_2}{r_2} + \frac{kQ_3}{r_3} + \ldots + \frac{kQ_n}{r_n} $
Potential due to linear charge distribution:
- $ dV = \frac{k\lambda(x)}{r}dx $
Potential due to surface charge distribution:
- $ dV = \frac{k\sigma(x, y)}{r}dA $
Potential due to volume charge distribution:
- $ dV = \frac{k\rho(x, y, z)}{r}d\tau $

Slide 25

Potential due to electric field:
- Electric field ( $ E $ ) at a point is the force experienced by a unit positive test charge
- Potential at a point is the work done to bring a unit positive test charge from infinity against the electric field
Electron Volt (eV):
- Unit of energy equal to the charge of an electron ( $ 1.6 \times 10^{-19} $ C) multiplied by one volt
Example: Ionization Energy of Hydrogen:
- Ionization energy = 13.6 eV
- Represents the energy required to remove an electron from a hydrogen atom
Calculating Energy using eV:
- Energy ( $ E $ ) = charge ( $ Q $ ) $ \times $ potential difference ( $ V $ )
- For example, if charge = 2 eV and potential difference = 5 V, then energy ( $ E $ ) = 2 $ \times $ 5 eV

Slide 26

Summary of Potential Due to Different Charge Distributions:
Point charges:
- Potential given by Coulomb’s law: $ V = \frac{kQ}{r} $
Multiple charges:
- Use the principle of superposition: $ V = \frac{kQ_1}{r_1} + \frac{kQ_2}{r_2} + \frac{kQ_3}{r_3} + \ldots + \frac{kQ_n}{r_n} $
Linear charge distribution:
- Potential $ dV = \frac{k\lambda(x)}{r}dx $
- Integrate to find the potential due to the entire distribution

Slide 27

Summary of Potential Due to Different Charge Distributions (contd):
Surface charge distribution:
- Potential $ dV = \frac{k\sigma(x, y)}{r}dA $
- Integrate to find the potential due to the entire surface
Volume charge distribution:
- Potential $ dV = \frac{k\rho(x, y, z)}{r}d\tau $
- Integrate to find the potential due to the entire volume
Potential due to electric field:
- Potential is the work done to bring a unit positive test charge from infinity against the electric field

Slide 28

Summary of Potential Due to Different Charge Distributions (contd):
Electron Volt (eV):
- Unit of energy equal to the charge of an electron ( $ 1.6 \times 10^{-19} $ C) multiplied by one volt
Applications of eV:
- Used to measure energies on an atomic and subatomic scale
- Quantizes energy levels and facilitates understanding of ionization, excitation, etc.
Calculating energy using eV:
- Energy ( $ E $ ) = charge ( $ Q $ ) $ \times $ potential difference ( $ V $ )

Slide 29

Example:
- Charge = 2 eV
- Potential difference = 5 V
- Energy ( $ E $ ) = 2 $ \times $ 5 eV
- Energy ( $ E $ ) = 10 eV
In summary, in this lecture, we discussed:
- The potential due to different charge distributions
- The concept of the electron volt as a unit of energy
- How to calculate the potential due to point charges, linear charge distributions, surface charge distributions, and volume charge distributions
- We also explored the relationship between potential, charge, and potential difference

Slide 30

Next Steps:
- Solve numerical problems related to potential due to different charge distributions and electron volts
- Practice applying these concepts to real-world scenarios
- Review the material to solidify your understanding
- Stay tuned for the next lecture!
Thank you for your attention!
Any questions?