Electric field of a uniformly charged disk

• member 392791
In summary, the conversation discusses finding the charge of a uniform disk with a continuous charge on the surface by using the area of concentric rings. The area formula used is σ(2πr dr), which is an approximation that becomes more accurate as dr becomes smaller. This approximation was not explained in the book, causing confusion, but the conversation clears it up.
member 392791
Hello,

I am looking at an example of finding the charge of a uniform disk with a continuous charge on the surface.

They go about the problem by finding the infinitesimal charge of concentric rings

dq = σdA = σ(2πr dr)

The part I don't understand is that they use the area as 2πr dr? The area of a ring would be ∏(R2^2 - R1^2), right?

Woopydalan said:
The area of a ring would be ∏(R2^2 - R1^2), right?

Correct, that's the exact area of the ring. However, if dr = R2 - R1 is small, the other formula is a very good approximation, which gets better and better as dr becomes smaller and smaller. To see this, let R2 = R1 + dr in the equation above, cancel out whatever you can, and then drop any terms with (dr)^2. Those terms become negligible compared to terms with just dr, when dr is very small.

Ok I just calculated it and now I see. The book didn't even bother to explain that approximation, which left me confused. Thank you for clearing that up so quickly =)

1. What is an electric field?

An electric field is a physical field that surrounds electrically charged particles or objects. It is a vector field, meaning it has both magnitude and direction, and is responsible for the electric force experienced by other charged particles in its vicinity.

2. How is the electric field of a uniformly charged disk calculated?

The electric field of a uniformly charged disk can be calculated using the formula E = σ/2ε0, where σ is the surface charge density of the disk and ε0 is the permittivity of free space.

3. What is the direction of the electric field of a uniformly charged disk?

The electric field of a uniformly charged disk is perpendicular to the surface of the disk at every point. This means the direction of the electric field lines will be radially outward from the center of the disk.

4. How does the electric field vary with distance from the center of the disk?

The electric field strength decreases as you move further away from the center of the disk. This is because the electric field follows an inverse square law, meaning it decreases with the square of the distance from the source.

5. Can the electric field of a uniformly charged disk be zero?

Yes, the electric field of a uniformly charged disk can be zero at the center of the disk. This is because the electric field lines from opposite sides of the disk cancel each other out at this point, resulting in a net electric field of zero.

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