few minutes left for my E&M homework

[SirStev0]

Equations 22-8 and 22-9 are approximations of the magnitude of the electric field of an electric dipole, at points along the dipole axis. Consider a point P on that axis at distance z = 6.00d from the dipole center(where d is the separation distance between the particles of the dipole). Let Eappr be the magnitude of the field at point P as approximated by Equations 22-8 and 22-9. Let Eact be the actual magnitude. By how much is the ratio Eappr/Eact less than 1?
wrong check mark

EDIT:

22-8
E = qd/2(pi)(primittivityconstant)(z^3)

22-9
E = P/2(pi)(primittivityconstant)(z^3)

 

[SirStev0]

anyone??? i have 10 minutes...

ohh and 22-9 and 22-8 are the equations for finding E of a point for a dipole

 

[SirStev0]

in case anyone was interested in the answer...

.013... thats right... .013

 

[SirStev0]

anyone think they can explain this at all?

 

[chuckywang]

Maybe you should tell us what Equations 22-8 and 22-9 are.

 

[SirStev0]

22-8
E = qd/2(pi)(primittivityconstant)(z^3)

22-9
E = P/2(pi)(primittivityconstant)(z^3)

 

[PottedMeat]

wait so you are given an E field description ( 22-8 and 9 ) on the axis of a dipole? are you supposed to find the value of the charges then just sum up the effects of each at point P? Wouldnt that just give you Eappr = Eact? Hmm don't remember calculating Apparent E fields in E&M

edit:

so whats 'P' - do you mean rho? like a linear charge distribution?

primittivity constant = permittivity = epsilon_0?

q = electron charge right?

d = ?

 

[SirStev0]

Originally posted by: PottedMeat
wait so you are given an E field description ( 22-8 and 9 ) on the axis of a dipole? are you supposed to find the value of the charges then just sum up the effects of each at point P? Wouldnt that just give you Eappr = Eact? Hmm don't remember calculating Apparent E fields in E&M

edit:

so whats 'P' - do you mean rho? like a linear charge distribution?

primittivity constant = permittivity = epsilon_0?

q = electron charge right?

d = ?

that's the long and the short of it. d is mentioned in the problem. My EE roommate couldnt figure out what exactly they were looking for and no Eact =/= Eappr

 

[Todd33]

Do your own homework!

 

[chuckywang]

I don't think the "d" in the equation is the same "d" in the problem. Can you elaborate on what the "d" in the equation is?

 

[Born2bwire]

Originally posted by: chuckywang
I don't think the "d" in the equation is the same "d" in the problem. Can you elaborate on what the "d" in the equation is?

d appears to be the distance of separation between the physical charges that make up the dipole. The P in his equation should be the dipole moment, which for an electric dipole is q*d. The field for a dipole moment is given by:
\vec{E} = p/(4\pi\epsilon_0r^3)(2\cos(\theta)\vec{r}+\sin(\theta)\vec{theta})
which reduces to his equation 22-8 when you substitute in theta = 0 to find the z directed field along the z axis.

A dipole, or any multipole, is the superposition of multiple static charges (at least in the case of our electrostatic problem). So the exact field is simply the summation of the electric fields resulting from your sources. In this case it is the two fields that arise from the positive and negative charges that make up your dipole. So it is as simple as just finding the coulombic field that arises from your static sources.

For small separation d's, the far-field case of a multipole can be very closely approximated using the dipole moment, that is, you degenerate the case into a single point source as opposed to the superposition of multiple point sources. In this case, you see that even at the very close distance of 6 times the separation of the charges, you only have a percent error of 1.3%. In addition, we see that we can better approximate the behavior of a physical dipole using a pure dipole by either reducing the separation of the source charges or by increasing the distance of the observer.

 

[chuckywang]

Originally posted by: Born2bwire

Originally posted by: chuckywang
I don't think the "d" in the equation is the same "d" in the problem. Can you elaborate on what the "d" in the equation is?

d appears to be the distance of separation between the physical charges that make up the dipole. The P in his equation should be the dipole moment, which for an electric dipole is q*d. The field for a dipole moment is given by:
\vec{E} = p/(4\pi\epsilon_0r^3)(2\cos(\theta)\vec{r}+\sin(\theta)\vec{theta})
which reduces to his equation 22-8 when you substitute in theta = 0 to find the z directed field along the z axis.

A dipole, or any multipole, is the superposition of multiple static charges (at least in the case of our electrostatic problem). So the exact field is simply the summation of the electric fields resulting from your sources. In this case it is the two fields that arise from the positive and negative charges that make up your dipole. So it is as simple as just finding the coulombic field that arises from your static sources.

For small separation d's, the far-field case of a multipole can be very closely approximated using the dipole moment, that is, you degenerate the case into a single point source as opposed to the superposition of multiple point sources. In this case, you see that even at the very close distance of 6 times the separation of the charges, you only have a percent error of 1.3%. In addition, we see that we can better approximate the behavior of a physical dipole using a pure dipole by either reducing the separation of the source charges or by increasing the distance of the observer.

The "d" in equation 22-8 seems to be half the distance between the charges. Otherwise, that equation is not right.

 

[Born2bwire]

Originally posted by: chuckywang

Originally posted by: Born2bwire

Originally posted by: chuckywang
I don't think the "d" in the equation is the same "d" in the problem. Can you elaborate on what the "d" in the equation is?

d appears to be the distance of separation between the physical charges that make up the dipole. The P in his equation should be the dipole moment, which for an electric dipole is q*d. The field for a dipole moment is given by:
\vec{E} = p/(4\pi\epsilon_0r^3)(2\cos(\theta)\vec{r}+\sin(\theta)\vec{theta})
which reduces to his equation 22-8 when you substitute in theta = 0 to find the z directed field along the z axis.

A dipole, or any multipole, is the superposition of multiple static charges (at least in the case of our electrostatic problem). So the exact field is simply the summation of the electric fields resulting from your sources. In this case it is the two fields that arise from the positive and negative charges that make up your dipole. So it is as simple as just finding the coulombic field that arises from your static sources.

For small separation d's, the far-field case of a multipole can be very closely approximated using the dipole moment, that is, you degenerate the case into a single point source as opposed to the superposition of multiple point sources. In this case, you see that even at the very close distance of 6 times the separation of the charges, you only have a percent error of 1.3%. In addition, we see that we can better approximate the behavior of a physical dipole using a pure dipole by either reducing the separation of the source charges or by increasing the distance of the observer.

The "d" in equation 22-8 seems to be half the distance between the charges. Otherwise, that equation is not right.

I haven't bothered to work out the actual numerical result, but in regards to just equation 22-8, d must mean the distance between the physical dipole. It works out fine. A factor of two is pulled out to the cosine term when we took the gradient of the potential (which falls off as 1/r^2, hence E~2/r^3).

EDIT: He is missing a few parentheses on his equations, the denominator is 2\pi\epsilon_0, not just 2.

 

[chuckywang]

Originally posted by: Born2bwire

Originally posted by: chuckywang

Originally posted by: Born2bwire

Originally posted by: chuckywang
I don't think the "d" in the equation is the same "d" in the problem. Can you elaborate on what the "d" in the equation is?

d appears to be the distance of separation between the physical charges that make up the dipole. The P in his equation should be the dipole moment, which for an electric dipole is q*d. The field for a dipole moment is given by:
\vec{E} = p/(4\pi\epsilon_0r^3)(2\cos(\theta)\vec{r}+\sin(\theta)\vec{theta})
which reduces to his equation 22-8 when you substitute in theta = 0 to find the z directed field along the z axis.

A dipole, or any multipole, is the superposition of multiple static charges (at least in the case of our electrostatic problem). So the exact field is simply the summation of the electric fields resulting from your sources. In this case it is the two fields that arise from the positive and negative charges that make up your dipole. So it is as simple as just finding the coulombic field that arises from your static sources.

For small separation d's, the far-field case of a multipole can be very closely approximated using the dipole moment, that is, you degenerate the case into a single point source as opposed to the superposition of multiple point sources. In this case, you see that even at the very close distance of 6 times the separation of the charges, you only have a percent error of 1.3%. In addition, we see that we can better approximate the behavior of a physical dipole using a pure dipole by either reducing the separation of the source charges or by increasing the distance of the observer.

The "d" in equation 22-8 seems to be half the distance between the charges. Otherwise, that equation is not right.

I haven't bothered to work out the actual numerical result, but in regards to just equation 22-8, d must mean the distance between the physical dipole. It works out fine. A factor of two is pulled out to the cosine term when we took the gradient of the potential (which falls off as 1/r^2, hence E~2/r^3).

EDIT: He is missing a few parentheses on his equations, the denominator is 2\pi\epsilon_0, not just 2.

The denominator should be 4*pi*epsilon_0.

 

[Born2bwire]

Originally posted by: chuckywang

Originally posted by: Born2bwire

Originally posted by: chuckywang

Originally posted by: Born2bwire

Originally posted by: chuckywang
I don't think the "d" in the equation is the same "d" in the problem. Can you elaborate on what the "d" in the equation is?

d appears to be the distance of separation between the physical charges that make up the dipole. The P in his equation should be the dipole moment, which for an electric dipole is q*d. The field for a dipole moment is given by:
\vec{E} = p/(4\pi\epsilon_0r^3)(2\cos(\theta)\vec{r}+\sin(\theta)\vec{theta})
which reduces to his equation 22-8 when you substitute in theta = 0 to find the z directed field along the z axis.

A dipole, or any multipole, is the superposition of multiple static charges (at least in the case of our electrostatic problem). So the exact field is simply the summation of the electric fields resulting from your sources. In this case it is the two fields that arise from the positive and negative charges that make up your dipole. So it is as simple as just finding the coulombic field that arises from your static sources.

For small separation d's, the far-field case of a multipole can be very closely approximated using the dipole moment, that is, you degenerate the case into a single point source as opposed to the superposition of multiple point sources. In this case, you see that even at the very close distance of 6 times the separation of the charges, you only have a percent error of 1.3%. In addition, we see that we can better approximate the behavior of a physical dipole using a pure dipole by either reducing the separation of the source charges or by increasing the distance of the observer.

The "d" in equation 22-8 seems to be half the distance between the charges. Otherwise, that equation is not right.

I haven't bothered to work out the actual numerical result, but in regards to just equation 22-8, d must mean the distance between the physical dipole. It works out fine. A factor of two is pulled out to the cosine term when we took the gradient of the potential (which falls off as 1/r^2, hence E~2/r^3).

EDIT: He is missing a few parentheses on his equations, the denominator is 2\pi\epsilon_0, not just 2.

The denominator should be 4*pi*epsilon_0.

No, it should be 2\pi\epsilon_0. Like I said, the solution is derived from the potential, which has a 1/r^2 dependence. When we take the gradient to get the electric field, the 1/r^2 brings out a multiplier of 2, so -2/r^3. This is done when we take the derivative with respect to the vector r hat. And then by choosing our observation point to be along the z-axis, we have only a field contribution from the r hat component and thus the 2 is pulled out making the denominator a factor of 2. This is shown in the general E field from a pure dipole that I had up in one of my earlier posts.

Another reason why it should be \frac{1}{2\pi\epsilon_0} is that we are dealing with the summation of two charges. The charges are close enough, and we are far enough away, that their field contributions are going to be approximately the same. If we take the observation along the z-axis, then the fields of each of the charges are going to be working in concert with eachother and we can see that we should have something similar to a factor of 2.

Or, in the wikipedia article, Text, they have a more generic vector form of the field of an electric dipole, but we can see that when the position and dipole vectors line up, we bring out a factor of 2.