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Potential due to electric dipole

by indie452
Tags: dipole, electric, potential
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indie452
#1
Apr17-10, 07:20 AM
P: 124
potential due to electric dipole is

V = p.r / 4(pi)(epsilon)r3

show the tangential component of the electrical field is

= psin(theta) / 4(pi)(epsilon)r3


what ive tried:

i assumed there are 2 charges separated by distance r
potential difference U = qV = -qE.r

so can i say E[tangential] = V/r ?
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Doc Al
#2
Apr17-10, 07:24 AM
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How is electric field related to potential?
indie452
#3
Apr17-10, 07:30 AM
P: 124
well E = -(grad)V
so could if i use d/dr? and get E = = -psin(theta) / 4(pi)(epsilon)r3

would i then just take E[tan] = Esin(theta)

if we assume the dipole is at an angle theta to the E field?

but what about the - sign

Doc Al
#4
Apr17-10, 08:16 AM
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Potential due to electric dipole

Quote Quote by indie452 View Post
well E = -(grad)V
Right.
so could if i use d/dr?
You want the tangential component.

Hint: [tex]\vec{p}\cdot \vec{r} = pr \cos\theta[/tex]

How do you take the derivative in the tangential direction?
indie452
#5
Apr17-10, 08:55 AM
P: 124
i'm not sure on how to do a tangential derivative...i may have done it in maths but didn't know thats what it was called.
is it related to spherical coords?
Doc Al
#6
Apr17-10, 08:59 AM
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Quote Quote by indie452 View Post
i'm not sure on how to do a tangential derivative...i may have done it in maths but didn't know thats what it was called.
is it related to spherical coords?
Yes. Look here: http://en.wikipedia.org/wiki/Gradien...dinate_Systems
indie452
#7
Apr17-10, 09:34 AM
P: 124
okay s if i use 1/r*d/dtheta

then Etan becomes

= -(-psin(theta) / 4(pi)(epsilon)r3)

okay i get that now...thanks

btw just so i can get the physical pic in my head, is the tangental component of V perpendicular to the moment 'bar', or is it simply horizontal?

cause i pictured the diagram to be as seen in attached so i'm not sure how it is prcos if not horizontal
Attached Thumbnails
qu8.jpg  
Doc Al
#8
Apr17-10, 09:46 AM
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Imagine that the dipole moment defines the z-axis. The field at any point (specified by the position vector r) will have components in all directions. The one we want is the tangential component, perpendicular to the position vector at any point. (In spherical coordinates, it will be the theta component.)

Note that pr cosθ is just the magnitude of the dot product of the vectors p and r that appears in the potential.
indie452
#9
Apr17-10, 09:55 AM
P: 124
Quote Quote by Doc Al View Post
Note that pr cosθ is just the magnitude of the dot product of the vectors p and r that appears in the potential.
oh yeah... of courseshould have realised that earlier...

thanks for the help, thats another revision topic i can tick off

btw one quick question unrelated - i'm doing a question at the moment that simply states that there is a laser beam with power = 15MW/m^2 and i need to give the peak amplitude of the electric field n the beam...
this question says its worth 10marks so the answer can't simply be power is proportional to the amplitude^2 can it?
Doc Al
#10
Apr17-10, 10:01 AM
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Quote Quote by indie452 View Post
btw one quick question unrelated - i'm doing a question at the moment that simply states that there is a laser beam with power = 15MW/m^2 and i need to give the peak amplitude of the electric field n the beam...
this question says its worth 10marks so the answer can't simply be power is proportional to the amplitude^2 can it?
I assume that that's the average intensity of the beam. Sure it's proportional to the the amplitude of the E field, but what's the proportionality constant? Hint: Review the Poynting vector.


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