- Recap of the electric field due to a point charge
- Introduction to electric dipoles
- Definition and properties of electric dipoles
- Calculation of electric field due to an electric dipole
- Derivation of the formula
- Explanation of the variables and their significance
- Example problem: Calculate the electric field at a point on the axial line of a dipole
- Comparison of field due to a point charge and an electric dipole
- Similarities and differences
- Visualization of field lines
- Electric field due to continuous charge distribution
- Calculating electric field using integration
- Example problem: Calculate the electric field due to a uniformly charged rod at a point on its axis
- Electric field due to a line of charge
- Derivation of the formula
- Example problem: Calculate the electric field at a distance from an infinitely long uniformly charged line
- Electric field due to a charged disk
- Derivation of the formula
- Example problem: Calculate the electric field at a point on the axis of a uniformly charged disk
- Electric field due to a charged spherical shell
- Derivation of the formula
- Example problem: Calculate the electric field at a point inside and outside a uniformly charged spherical shell
Slide 11
Electric Field due to a Charged Ring
- Consider a circular ring with radius R and charge Q
- The electric field at a point on the axis of the ring is given by:
$$E = \frac{kQz}{(z^2 + R^2)^{\frac{3}{2}}}$$
- Where:
- k is the Coulomb’s constant
- Q is the charge of the ring
- z is the distance along the axis from the center of the ring
Example problem: Calculate the electric field at a point P on the axis of a uniformly charged ring with a radius of 2 cm and a charge of 20 μC. The distance of point P from the center of the ring is 5 cm.
- Solution:
- Given: R = 2 cm, Q = 20 μC, z = 5 cm
- Plug in the values into the formula:
- $E = \frac{(9 \times 10^9 \times 20 \times 10^{-6} \times 5)}{(5^2 + 2^2)^{\frac{3}{2}}}$
- Calculate the final result.
Slide 12
Electric Field due to a Spherical Shell
- Consider a uniformly charged spherical shell with radius R and charge Q
- The electric field at a point outside the shell is given by:
$$E = \frac{kQ}{r^2}$$
- Where:
- k is the Coulomb’s constant
- Q is the charge of the shell
- r is the distance from the center of the shell to the point
Example problem: Calculate the electric field at a point outside a uniformly charged spherical shell with a charge of 5 μC and a radius of 3 cm. The distance of the point from the center of the shell is 4 cm.
- Solution:
- Given: Q = 5 μC, R = 3 cm, r = 4 cm
- Plug in the values into the formula:
- $E = \frac{(9 \times 10^9 \times 5 \times 10^{-6})}{(0.04)^2}$
- Calculate the final result.
Slide 13
Electric Field due to a Non-uniformly Charged Line
- In some cases, the charge distribution along a line may not be uniform
- To calculate the electric field, we divide the line into small segments and sum up the contributions from each segment
$$dE = \frac{kQ}{r^2}cos\theta$$
- Example problem: Consider a non-uniformly charged line segment with a total charge of 10 μC. The charge distribution is given by λ = 2xz, where x is the position along the line. Calculate the electric field at a point P located at a distance of 3 cm from the line segment.
- Solution:
- Divide the line segment into small segments and calculate the charge on each segment using the given charge distribution λ = 2xz
- Use the formula $dE = \frac{kQ}{r^2}cos\theta$ to calculate the electric field contribution from each segment
- Sum up the contributions from all segments to get the total electric field at point P
Slide 14
Electric Field due to a Non-uniformly Charged Ring
- Similar to a non-uniformly charged line, a non-uniformly charged ring can also have a non-uniform charge distribution
- We can calculate the electric field at a point P using the same principle as before, dividing the ring into small segments and summing up the contributions from each segment
- Example problem: Consider a non-uniformly charged ring with a total charge of 10 μC. The charge distribution is given by λ = R^2sin^2θ, where R is the radius of the ring and θ is the angle measured from the positive x-axis. Calculate the electric field at a point P located at a distance of 5 cm along the axis of the ring.
- Solution:
- Divide the ring into small segments and calculate the charge on each segment using the given charge distribution λ = R^2sin^2θ
- Use the formula $dE = \frac{kQ}{r^2}cos\theta$ to calculate the electric field contribution from each segment
- Sum up the contributions from all segments to get the total electric field at point P
Slide 15
Electric Field due to a Non-uniformly Charged Disk
- A non-uniformly charged disk also requires a similar approach to calculate the electric field at a point
- Divide the disk into small segments and sum up the contributions from each segment
Example problem: Consider a non-uniformly charged disk with a total charge of 20 μC. The charge distribution is given by σ = a(2 - x), where a is a constant and x is the position along the radial direction of the disk. Calculate the electric field at a point P located at a distance of 4 cm from the center of the disk along the axis.
- Solution:
- Divide the disk into small segments and calculate the charge on each segment using the given charge distribution σ = a(2 - x)
- Use the formula $dE = \frac{kQ}{r^2}cos\theta$ to calculate the electric field contribution from each segment
- Sum up the contributions from all segments to get the total electric field at point P
Slide 16
Electric Field due to a Non-uniformly Charged Sphere
- A non-uniformly charged sphere can also be analyzed using the same approach
- Divide the sphere into small segments and sum up the contributions from each segment
Example problem: Consider a non-uniformly charged sphere with a total charge of 15 μC. The charge distribution is given by ρ = \frac{(3r + 2)}{r^2}, where r is the distance from the center of the sphere. Calculate the electric field at a point P located at a distance of 5 cm from the center of the sphere.
- Solution:
- Divide the sphere into small segments and calculate the charge on each segment using the given charge distribution ρ = \frac{(3r + 2)}{r^2}
- Use the formula $dE = \frac{kQ}{r^2}$ to calculate the electric field contribution from each segment
- Sum up the contributions from all segments to get the total electric field at point P
Field Due To Dipole And Continuous Charge Distributions - Non-uniform electric field
Slide 21
Electric Field due to a Non-uniformly Charged Cylinder
- A non-uniformly charged cylinder can also be analyzed using a similar approach as before
- Divide the cylinder into small segments and sum up the contributions from each segment
Example problem: Consider a non-uniformly charged cylinder with a total charge of 10 μC. The charge distribution is given by σ = a(2 - z), where a is a constant and z is the position along the axis of the cylinder. Calculate the electric field at a point P located at a distance of 3 cm from the center of the cylinder along its axis.
- Solution:
- Divide the cylinder into small segments and calculate the charge on each segment using the given charge distribution σ = a(2 - z)
- Use the formula $dE = \frac{kQ}{r^2}cos\theta$ to calculate the electric field contribution from each segment
- Sum up the contributions from all segments to get the total electric field at point P
Slide 22
Electric Field due to a Non-uniformly Charged Plane
- A non-uniformly charged plane can be analyzed using the same principle
- Divide the plane into small segments and sum up the contributions from each segment
Example problem: Consider a non-uniformly charged plane with a total charge of 15 μC. The charge distribution is given by σ = a(x + y), where a is a constant, x and y are the coordinates on the plane. Calculate the electric field at a point P located at a distance of 2 cm from the plane.
- Solution:
- Divide the plane into small segments and calculate the charge on each segment using the given charge distribution σ = a(x + y)
- Use the formula $dE = \frac{kQ}{r^2}cos\theta$ to calculate the electric field contribution from each segment
- Sum up the contributions from all segments to get the total electric field at point P
Slide 23
Electric Field due to a Non-uniformly Charged Sphere Shell
- A non-uniformly charged spherical shell requires a similar approach for analysis
- Divide the shell into small segments and sum up the contributions from each segment
Example problem: Consider a non-uniformly charged spherical shell with a total charge of 20 μC. The charge distribution is given by σ = a(3 - r), where a is a constant and r is the distance from the center of the shell. Calculate the electric field at a point P located at a distance of 6 cm from the center of the shell.
- Solution:
- Divide the shell into small segments and calculate the charge on each segment using the given charge distribution σ = a(3 - r)
- Use the formula $dE = \frac{kQ}{r^2}$ to calculate the electric field contribution from each segment
- Sum up the contributions from all segments to get the total electric field at point P
Slide 24
Electric Field Due to Continuous Charge Distributions - Line Integral
- For more complex charge distributions, we can use the concept of line integrals
- Line integrals help us to calculate the electric field due to continuous charge distributions
Example problem: Consider a continuous charge distribution along a curve C. Calculate the electric field at a point P on curve C using the line integral method.
- Solution:
- Express the continuous charge distribution as a function of position on the curve C
- Write down the line integral expression for the electric field
- Evaluate the line integral to calculate the electric field at point P
Slide 25
Example: Electric Field due to a Spiral Wire
- Consider a wire formed in the shape of a spiral with a uniform charge distribution
- The charge density is given by λ
Example problem: Calculate the electric field at a point P located at a distance of 5 cm above the center of the spiral wire.
- Solution:
- Express the charge density λ as a function of position along the wire
- Write down the line integral expression for the electric field due to a spiral wire
- Evaluate the line integral to calculate the electric field at point P
Slide 26
Example: Electric Field due to a Helical Wire
- Consider a wire formed in the shape of a helix with a uniform charge distribution
- The charge density is given by λ
Example problem: Calculate the electric field at a point P located at a distance of 4 cm above the center of the helical wire.
- Solution:
- Express the charge density λ as a function of position along the wire
- Write down the line integral expression for the electric field due to a helical wire
- Evaluate the line integral to calculate the electric field at point P
Slide 27
Electric Field due to Continuous Charge Distributions - Surface Integral
- For charge distributions on surfaces, we can use the concept of surface integrals
- Surface integrals help us to calculate the electric field due to continuous charge distributions
Example problem: Consider a continuous charge distribution on a surface S. Calculate the electric field at a point P outside the surface using the surface integral method.
- Solution:
- Express the charge distribution as a function of position on the surface S
- Write down the surface integral expression for the electric field
- Evaluate the surface integral to calculate the electric field at point P
Slide 28
Example: Electric Field due to a Charged Sphere
- Consider a uniformly charged solid sphere of radius R with charge density ρ
Example problem: Calculate the electric field at a point P located at a distance of 6 cm from the center of the charged sphere.
- Solution:
- Express the charge density ρ as a function of position within the sphere
- Write down the surface integral expression for the electric field due to a charged sphere
- Evaluate the surface integral to calculate the electric field at point P
Slide 29
Example: Electric Field due to a Charged Cylinder
- Consider a uniformly charged infinite cylinder with charge density σ
Example problem: Calculate the electric field at a point P located at a distance of 5 cm from the axis of the cylinder.
- Solution:
- Express the charge density σ as a function of position within the cylinder
- Write down the surface integral expression for the electric field due to a charged cylinder
- Evaluate the surface integral to calculate the electric field at point P
Slide 30
Summary
- Recap of the electric field due to a non-uniform charge distribution
- Line and surface integrals for calculating the electric field
- Examples of electric fields due to various continuous charge distributions