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If I am right you can only do vector product for vectors. Here you have dipole moment $= q×d$. Then are we assuming charge is a vector? But how then does charge obey the triangle law of vector addition (,ie apart from having a direction a quantity must also obey the vector addition law). This is the reason why current isn't a vector. But I don't understand how we assume charge is a vector?

Another confusion I stumbled upon is, the direction of a vector product should be perpendicular to both the vectors. In this case, dipole moment is from negative to positive. Which is along the direction of the distance vector (distance vector? You mean position vector???)

I think I am missing out two things. How that vector product is even possible (having rejected the hypothesis of charge being a vector). And two what is the distance vector? What's it's direction. In nutshell, how do you describe both the "vectors"

Thanks in advance. Apologies if I made a typo. I am on mobile right now, leaving my lunch cold and annoyed that this formula $q×d$ even exists .

Edit: here's the picture since I feel it tells how I got confused

![enter image description here](https://i.sstatic.net/hDeqS.jpg)

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    $\begingroup$ Well I think I'll just send a pic of my textbook so future readers will know where I got my confusion from $\endgroup$
    – Adil Mohammed
    Commented May 15, 2021 at 13:16
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    $\begingroup$ Please do not post images of texts you want to quote, but type it out instead so it is readable for all users and so that it can be indexed by search engines. For formulae, use MathJax instead. (Also, if you absolutely must post a picture, at least rotate it such that it's oriented in the same way as the rest of the text) $\endgroup$
    – ACuriousMind ♦
    Commented May 15, 2021 at 13:19
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    $\begingroup$ yes, that's a bad notation. Normally, we don't use such notation. $\endgroup$
    – mathLover
    Commented May 15, 2021 at 13:19
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    $\begingroup$ The symbol $\times$ in your book means multiplication and not cross product. The book formats vectors in bold face, so you can also see how the $\times$ is not written between two vectors but between scalars. $\endgroup$
    – Steeven
    Commented May 15, 2021 at 13:20
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    $\begingroup$ Evidently this textbook is one of the (too) many that are published without a physicist reviewing and editing it. I'm teaching from one now (not my choice). It looks like a physics textbook, but it isn't. Like those cheap writing tools given to many teachers. They look like pencils, but they can't be sharpened, and they wear down quickly, Good for one use. [end rant] $\endgroup$
    – garyp
    Commented May 15, 2021 at 13:50

2 Answers 2

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No, charge is not a vector quantity. Your definition of dipole moment is incorrect. The correct definition is $$\mathbf p=q\mathbf d$$ where $\mathbf d$ is the separation between negative and positive $q$ charge, directed from the negative towards the positive charge. $\mathbf d$ is the vector.

Edit: Your book is not talking about a vector cross product. Look at the notation. The vectors have been typed in boldface. The book is simply multiplying a scalar, charge, with the vector $\mathbf d$ I mentioned.

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No. The electric charge is not a vector.

The electric dipole moment has a magnitude as

$$p=qd,$$

where $d$ is the distance between charges. Since the dipole moment is useful if defined as a quantity with magnitude and direction it can be expressed in vector form by introducing $\vec d$ as the displacement vector (directed from the negative charge to the positive charge). Thus in textbooks it's defined as $$\vec p= q \vec d.$$

Hope this helps.

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