Magnitude of an Electric Field due to a dipole

In summary: This caused me to incorrectly calculate the magnitude of the electric field.Oh my god I see what I did wrong. The question says ##\langle 0, 2\times 10^{-8}, 0\rangle## and I misread it as ##\langle 0.2\times 10^{-8}, 0\rangle##. This caused me to incorrectly calculate the magnitude of the electric field.
  • #1
Zack K
166
6

Homework Statement


A dipole is located at the origin, and is composed of charged particles with charge +e and -e, separated by a distance 2x10-10m along the x-axis.

Calculate the magnitude of the electric field due to the dipole at location ##\langle 0.2\times 10^{-8}, 0, 0\rangle##m

Homework Equations


##|\vec E_{axis}|=\frac {2kqs} {r^3}## for ##r\gg s##

The Attempt at a Solution


It seems simple and yet I have no idea what's wrong. I just plugged all the values in.
##|\vec E_{axis}|=\frac {2(9\times 10^9)(1.602\times 10^{-19})(2\times 10^{-10})} {(0.2\times 10^{-0.8})^3}=7.21\times 10^7N/C##. But the answer is ##3.6\times 10^4N/C##
I used this formula with the assumption that the r is much greater than s, which it is. But I even tried the formula where the difference between s and r is not a vast difference, yet I still got the wrong answer.
 
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  • #2
Zack K said:

Homework Equations


##|\vec E_{axis}|=\frac {2kqs} {r^3}## for ##r\gg s##
Where did you get this? (That 2 shouldn't be there.)
 
  • #3
Doc Al said:
Where did you get this? (That 2 shouldn't be there.)
From the textbook my class is using. Matter & Interactions 4th edition volume 2. I'm pretty sure the 2 is supposed to be there since this is a dipole and not a point charge. I can show you the derivation they did.
 
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  • #4
Zack K said:
From the textbook my class is using. Matter & Interactions 4th edition volume 2. I'm pretty sure the 2 is supposed to be there since this is a dipole and not a point charge. I can show you the derivation they did.
Oops, my bad. The formula is correct. I suspect the answer is a mistake -- they used the formula for the perpendicular axis instead of the parallel.
 
  • #5
Doc Al said:
Oops, my bad. The formula is correct. I suspect the answer is a mistake -- they used the formula for the perpendicular axis instead of the parallel.
It seems weird for an answer to be this wrong though. If I get rid of the 2 and scale it down by a factor of 1000 then I get the answer right. There's a second part to the question where again, to get the answer right I have to scale it down by a factor of a thousand. Isn't it bizarre for a published book on physics used in a class to miss something like that?
 
  • #6
Ah, see checkpoint 5 in Chapter 13.6. (I found the book.) They ask two questions, perhaps you mixed up the answers.

The distance is 2 x 10-8, not 0.2 x 10-8.
 
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  • #7
Doc Al said:
Ah, see checkpoint 5 in Chapter 13.6. (I found the book.) They ask two questions, perhaps you mixed up the answers.
Oh my god I see what I did wrong. The question says ##\langle 0, 2\times 10^{-8}, 0\rangle## and I misread it as ##\langle 0.2\times 10^{-8}, 0\rangle##
 

Related to Magnitude of an Electric Field due to a dipole

1. What is an electric dipole?

An electric dipole is a pair of equal and opposite charges separated by a small distance. It can be thought of as a positive charge and a negative charge that are close together but not touching.

2. How is the magnitude of an electric field due to a dipole calculated?

The magnitude of an electric field due to a dipole is calculated by taking the product of the charge magnitude and the distance between the two charges, and dividing it by the cube of the distance from the dipole to the point where the electric field is being measured.

3. What factors affect the magnitude of an electric field due to a dipole?

The magnitude of an electric field due to a dipole is affected by the distance between the charges, the magnitude of the charges, and the orientation of the dipole relative to the point where the electric field is being measured.

4. How does the magnitude of an electric field due to a dipole change with distance?

The magnitude of an electric field due to a dipole decreases with distance according to an inverse cube law. This means that as the distance from the dipole increases, the magnitude of the electric field decreases rapidly.

5. What is the direction of the electric field due to a dipole?

The direction of the electric field due to a dipole is always from the positive charge towards the negative charge. At points along the axis of the dipole, the electric field is directed towards the negative charge, while at points perpendicular to the axis, the electric field is directed towards the positive charge.

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