Comparing Problem #1 & #2: Why the Difference?

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Homework Help Overview

The discussion revolves around two physics problems involving electric fields generated by charged objects: a charged ring and a charged disk. Participants are exploring the differences in the integration approaches required for each problem.

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants are questioning why the first problem integrates over charge elements while the second problem requires limits of integration due to varying contributions from different elements. There is also discussion about the nature of the components being integrated in each case.

Discussion Status

Some participants have offered insights into the differences in the integration process between the two problems, noting that the field contributions vary in the second case. There is ongoing exploration of the implications of these differences, with no explicit consensus reached.

Contextual Notes

There are mentions of potential corrections regarding the expressions used in the integrals, indicating that participants are actively engaging with the mathematical details of the problems.

UrbanXrisis
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My question involves two example problems.

1. A ring of radius [tex]a[/tex] carries a uniforly distributed positive total charge [tex]Q[/tex]. Calculate the electrical field due to the ring at a point [tex]P[/tex] lying a distance [tex]x[/tex] from its center along the central axis perpendicular to the plane of the ring.

[tex]dE_x=dEcos \theta = (k \frac{dq}{r^2})\frac {x}{r} = \frac{kx}{(x^2+a^2)^{3/2}} dq[/tex]

[tex]E_x= \int \frac{kx}{(x^2+a^2)^{3/2}}dq=\frac{kx}{(x^2+a^2)^{3/2}} \int dq[/tex]

[tex]E_x= \frac{kx}{(x^2+a^2)^{3/2}}Q[/tex]

2. A disk of radius [tex]R[/tex] has a uniform surface charge density [tex]\sigma[/tex]. Calculate the electrical field at a point [tex]P[/tex] that lies along the central perpendicular axis of the disk and a distance [tex]x[/tex] from the center of the disk.

[tex]dq=2 \pi \sigma r dr[/tex]

[tex]dE=\frac{kx}{(x^2+a^2)^{3/2}}(2 \pi \sigma r dr)[/tex]

[tex]E=kx \pi \sigma \int_0 ^R \frac{2r dr}{(x^2+a^2)^{3/2}}[/tex]

My question is:
In problem #1 vs Problem #2, why are there limits on #2 and why does #1 only integrate the dq?
 
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maybe because in the first case the force contributed by each element is the same where as the in second case the force exerted by each ring of thickness dr is different.so u hv to integrate to get the total force by the disc(or all the rings of thickness dr)..hence the field...does this suffice??
 
well you've got the multiplication of two components, why would you integrate the finite component?
 
GCT said:
well you've got the multiplication of two components, why would you integrate the finite component?

that's what I'm wondering about too... there's:

[tex]E=kx \pi \sigma \int_0 ^R \frac{2r dr}{(x^2+a^2)^{3/2}}[/tex]

which I am integrating a finite componet of x and a while in #1, I don't integrate the [tex](x^2+a^2)^{3/2}[/tex]

why?
 
In #1 the field contribution from each element of charge (dq) is a constant, so it drops out of the integral. The limits of integration are implied to be from q = 0 to q = Q.

Not so for #2, since the field contribution from each ring of charge depends on the radius.

Realize that in the integral for #2, "a" should be replaced by the variable "r". "a" was a constant in #1, but is not in #2.
 
hey check ur working again...i think it is r^2 + x^2 in the denominator and not wht u hv written...
 
yes, you are correct
 

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