Question about the E-field along center axis of charged ring

In summary, the use of "dE" and "dQ" notation in this problem is not for derivatives, but rather for integration. It is a standard procedure to split the body into small elements and treat the measurements on them as constant. In this case, the charge is represented by dq and the total charge can be found by integrating dq. The Ex for each charge is calculated using kxdq/(x2+a2)3/2 and the total Ex is found by integrating this expression. This method makes it easier to solve for the total electric field in the given problem.
  • #1
Loopas
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http://www.physics.udel.edu/~watson/phys208/exercises/kevan/efield1.html

I'm having trouble understanding the derivative notation. I know that "dE" and dQ" have to do with the derivative, but what exactly does that mean in the context of this problem? For some reason I find it difficult to understand why this derivative notation is used in so many physics questions. How do I know what variables have derivative notation?
 
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  • #2
Hi Loopas! :smile:
Loopas said:
I'm having trouble understanding the derivative notation. I know that "dE" and dQ" have to do with the derivative, but what exactly does that mean in the context of this problem? … How do I know what variables have derivative notation?

It isn't a derivative … in fact, you're going to integrate it.

dq is the charge on a tiny piece of the ring.

This is a standard procedure: you split the body into slices that are so small that you can regard all measurements on them as constant.

(for example, if you know the density of a body as a function of position, you say consider a small element of volume dxdydz, it can be assumed to have a constant density ρ, so the mass is ρdxdydz, and the total mass is ∫ ρdxdydz)

In this case, the charge is dq, the total charge is ∫ dq, the Ex for that charge is kxdq/(x2+a2)3/2, and so the total Ex is ∫ kxdq/(x2+a2)3/2 :wink:


(btw, in this case, all measurements on the small charge really are constant, so ther's actaully no assumption here … but usually there is)
 

FAQ: Question about the E-field along center axis of charged ring

What is the E-field along the center axis of a charged ring?

The E-field along the center axis of a charged ring is a measure of the strength and direction of the electric field at various points along the axis, caused by the charged particles in the ring.

How is the E-field along the center axis of a charged ring calculated?

The E-field along the center axis of a charged ring can be calculated using the equation E = kqz/r^2, where k is the Coulomb constant, q is the charge of the ring, z is the distance from the center axis, and r is the radius of the ring.

Is the E-field along the center axis of a charged ring constant?

No, the E-field along the center axis of a charged ring is not constant. It varies depending on the distance from the ring and the charge distribution of the ring. The E-field is strongest at the center of the ring and decreases as the distance from the ring increases.

How does the E-field along the center axis of a charged ring differ from the E-field at other points?

The E-field along the center axis of a charged ring differs from the E-field at other points because it is directly affected by the charge distribution of the ring. At other points, the E-field may be affected by other nearby charged objects and the distance from those objects.

How does the E-field along the center axis of a charged ring relate to the electric potential?

The E-field along the center axis of a charged ring is directly related to the electric potential. The electric potential is the energy per unit charge at a point in the electric field, and it can be calculated using the equation V = kq/r, where k is the Coulomb constant, q is the charge of the ring, and r is the distance from the center axis.

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