Finding dipole moment vector / Electric Field problem

Jan 2016
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Numbers 3 and 4 from the included image are giving me some issues. Help?
 

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topsquark

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Apr 2008
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On the dance floor, baby!
p = q d

where d is the displacement between two equal charges. So look at a charge dq on one side of the circle. It has a "partner" on the other side so between them we have a dipole moment of dp = dq R, where R is the radius of the circle. So p = (integral) dp

How far have you been able to get with the other one?

-Dan
 

Pmb

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Apr 2009
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Boston's North Shore
p = q d

where d is the displacement between two equal charges.
The magnitude of the charges are the same but the signs are different. The direction of d is from the negative charge to the positive charge.
 
Jan 2016
2
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Would we use d = charge separation = 2R, not d=R?

So, p = ∫dp = ∫d(dq) = ∫(2R)dq = ∫(2R)[λRdθ]

Where λ=Q/s, so Q = λs, or dq=λds=λRdθ, resulting in...

So, p = 2λR^2∫dθ from pi to -pi?

This isn't feeling right since nothing is mentioned about charge density...??
 

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Last edited:

Pmb

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sebastian said:
Would we use d = charge separation = 2R, not d=R?
The value d = 2R would be used.

Sebastian said:
So, p = ∫dp = ∫d(dq) = ∫(2R)dq = ∫(2R)[λRdθ]

Where λ=Q/s, so Q = λs, or dq=λds=λRdθ, resulting in...

So, p = 2λR^2∫dθ from pi to -pi?
That is incorrect. The range of θ is from 0 to pi.
Sebastian said:
This isn't feeling right since nothing is mentioned about charge density...??
All of the information required to determine λ= linear charge density is available in the image. It's λ = Q/S = Q/pi*R. S = Half of the circumference of the ring.
 
Jan 2016
2
0
The value d = 2R would be used.


That is incorrect. The range of θ is from 0 to pi.

All of the information required to determine λ= linear charge density is available in the image. It's λ = Q/S = Q/pi*R. S = Half of the circumference of the ring.
Ah, yes. Thank you very much! I think I got it now.