Electric Field of a Charged Disk

In summary, the area of a ring section is determined by finding the difference between the areas of two concentric circles, one with a radius of a and the other with a radius of a + da. This results in an area of 2*pi*a*da, as da is considered to be an infinitesimal and its square is negligible. This can also be visualized as a rectangle with a length of 2*pi*a and a width of da, which again results in an area of 2*pi*a*da. By using an integral, we can find the exact area rather than just an approximate area.
  • #1
davezhan
3
0
I'm having a brain freeze and have a hard time understanding why the area of the ring is 2*pi*a*da. Can someone explain why it is the circumference times da?

Link to derivation: www.phys.uri.edu/~gerhard/PHY204/tsl36.pdf
 
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  • #2
davezhan said:
I'm having a brain freeze and have a hard time understanding why the area of the ring is 2*pi*a*da. Can someone explain why it is the circumference times da?

Link to derivation: www.phys.uri.edu/~gerhard/PHY204/tsl36.pdf

How would you write the equation for the area of that ring section? What happens when you simplify what you've written?
 
  • #4
Think of the ring as the region between two circles- one of radius a, the other of radius a+ da. The area of the inner circle is [itex]\pi a^2[/itex] and the area of the outer circle is [itex]\pi (a+ da)^2[/itex]. The area between them is [itex]\pi (a+da)^2- \pi a^2[/itex][itex]= \pi (a^2+ 2ada+ da^2)- \pi a^2[/itex][itex]= 2\pi a da+ \pi da^2[/itex]. Since da is an "infinitesmal", its square is negligible and the area is [itex]2\pi a da[/itex]. By saying that "da is an infinitesmal" I mean that this is true in the limit sense for very small da.

Here's another way to look at it: Imagine opening that strip up to a "rectangle". It's length is the circumference of the circle, [itex]2\pi a[/itex], and it's width is da. The area of that "rectangle" is "length times width", [itex]2\pi a da[/itex]. I have put "rectangle" in quotes because, of course, you cannot "open up" a circular strip into a rectangle. This is, again, only true in the limit sense.

If you were to take da to be any finite length, [itex]2\pi a da[/itex] would give you an approximate area, not an exact area. But you can use "da" in an integral to get the exact area.
 
  • #5
Thank you for your help! The above post helped to clarify things tremendously.
 

FAQ: Electric Field of a Charged Disk

What is an electric field?

The electric field is a physical quantity that describes the force that an electric charge experiences in a given space. It is a vector quantity, meaning it has both magnitude and direction.

What is a charged disk?

A charged disk is a two-dimensional object with a uniform distribution of electric charge on its surface. It can be formed by cutting a charged sphere in half, resulting in two charged disks with opposite charges on their surfaces.

How is the electric field of a charged disk calculated?

The electric field of a charged disk can be calculated using the formula E = σ/2ε₀, where σ is the surface charge density (charge per unit area) and ε₀ is the permittivity of free space. This formula assumes that the disk has a radius much larger than its thickness.

What is the direction of the electric field of a charged disk?

The electric field of a charged disk points radially outward from the center of the disk if the charge is positive, and radially inward if the charge is negative. This is because the electric field lines are perpendicular to the surface of the disk and point away from positive charges and towards negative charges.

How does the electric field of a charged disk change as distance from the disk increases?

The electric field of a charged disk decreases as distance from the disk increases. This is because the electric field follows an inverse square law, meaning it decreases with the square of the distance from the source charge. As you move further away from the disk, the electric field becomes weaker.

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