Electric field

  • #1
1,197
1
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?
 

Answers and Replies

  • #2
Wiz
21
0
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??
 
  • #3
GCT
Science Advisor
Homework Helper
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well you've got the multiplication of two components, why would you integrate the finite component?
 
  • #4
1,197
1
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 dont integrate the [tex](x^2+a^2)^{3/2}[/tex]

why?
 
  • #5
Doc Al
Mentor
44,986
1,254
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.
 
  • #6
Wiz
21
0
hey check ur working again...i think it is r^2 + x^2 in the denominator and not wht u hv written.....
 
  • #7
1,197
1
yes, you are correct
 

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