Magnitude of an Electric Field due to a dipole

[Zack K]

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.

 

[Doc Al]

Homework Equations

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

Where did you get this? (That 2 shouldn't be there.)

 

[Zack K]

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.

 

[Doc Al]

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.

 

[Zack K]

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?

 

[Doc Al]

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.

 

[Zack K]

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 $\langle 0.2\times 10^{-8}, 0\rangle$