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Magnitude and Direction of Electric Field - Question

  • Thread starter UNG
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  • #1
UNG
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I also have another question that I am very confused about

Here it is
4ques-1.jpg


I think you are suppose to find the magnitude and direction of the electric field and the formula needed is,

E = 1 / (4x 3.14 x e0) x (density x distance / R^2)

And I was told integration is needed too??

but I really dont know where to start, a correct formula to show the steps would be helpful.

Many thanks !
 

Answers and Replies

  • #2
Kurdt
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Yes integration will be required. First you will have to find the linear charge density of the lines of charge. This is fairly simple to do. Integration will be needed as you will have to sum the contributions of the infinitesimal charges of the ring. Have you seen any examples in any text books?
 
  • #3
UNG
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Um no I don really understand the examples in the textbooks.
 
  • #4
Kurdt
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Ok. Well how would you determine the arc length of a circle?
 
  • #5
UNG
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With the formula,

L = angle x 3.14 x R / 180
 
  • #6
learningphysics
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Get the net x-component of the field due to the negative segment... get the net y-component of the field due to the negative segment...

then do the same thing for the positive component.

only plug in the numbers at the end...

If theta is the angle as measured from the negative y-axis going clockwise... an infintesimal charge is [tex]dq = Rd\theta*\sigma[/tex] where [tex]\sigma[/tex] is the charge density. what is sigma in terms of q? what is the x-component of this field? what is the y-component of this field?
 
  • #7
UNG
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Could I also find the x -component from the formula,

[tex]\vec{E} = \frac{1}{4 \pi \epsilon _0} \int \frac{\lambda(\vec{r}^{\prime}) ds^{\prime}}{R^2}[/tex]
 
  • #8
learningphysics
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Could I also find the x -component from the formula,

[tex]\vec{E} = \frac{1}{4 \pi \epsilon _0} \int \frac{\lambda(\vec{r}^{\prime}) ds^{\prime}}{R^2}[/tex]
I don't think that would work exactly as it is... to get the x-component... you need to add up the x-components of each infinitesimal charge.

this would be better:

[tex]\vec{E} = \frac{1}{4 \pi \epsilon _0} \int \frac{\lambda(\vec{r}^{\prime}) ds^{\prime}}{R^2}cos(\theta)[/tex]
 
  • #9
UNG
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But there is no angle given to calculate it?
 
  • #10
learningphysics
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But there is no angle given to calculate it?
The angle changes as you integrate... you set up the angle yourself to integrate...
 
  • #11
UNG
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okay so that means you could choose your angle,

for instance 180 degrees and put it in the formula
 
  • #12
learningphysics
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No. suppose theta is the angle from the negative y-axis going clockwise... the charge density is sigma q/(0.25*2pi*r) = 2q/pi*r

an differential element has charge [tex]\sigma*ds = \sigma*rd\theta[/tex]

The magnitude of the field due to this charge is [tex]\frac{k\sigma*ds}{r^2}[/tex]. The x-component of this field is:

[tex]-\frac{k\sigma*ds}{r^2}sin(\theta)}[/tex]

which equals

[tex]-\frac{k\sigma*rd\theta}{r^2}sin(\theta)[/tex]...

which is

[tex]-\frac{k(2q/\pi r)*rd\theta}{r^2}sin(\theta)[/tex]...

which is:

[tex]\frac{-2kqd\theta}{\pi r^2}sin(\theta)[/tex]...


if you integrate this from theta equals 0 to pi/2... you get the x-component due to the -q part...

the y-component is [tex]-\frac{k\sigma*ds}{r^2}cos(\theta)}[/tex]

integrate this from 0 to pi/2 and you get the y-component due to the -q part...

do the same type of thing with the +q part.
 

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