Coulomb's law in vector form

logearav

Joined Aug 19, 2011
243
Revered Members,
Kindly see my attachment.
Why for F$$_{21}$$ the unit vector is written as r$$_{12}$$ instead of r$$_{21}$$?

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steveb

Joined Jul 3, 2008
2,436
Revered Members,
Kindly see my attachment.
Why for F$$_{21}$$ the unit vector is written as r$$_{12}$$ instead of r$$_{21}$$?
It simply eliminates the need for a negative sign in the formula. Since this is a vector equation, the equation must give the correct direction for the force vector. This means a negative sign is needed. But swapping the indices from 21 to 12 is the same as a negative sign.

logearav

Joined Aug 19, 2011
243
Thanks for the reply steveb. Now look at this attachment. This material uses r$$_{21}$$ for F$$_{21}$$ which contradicts my previous attachment. Thats why i am confused.

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steveb

Joined Jul 3, 2008
2,436
Thanks for the reply steveb. Now look at this attachment. This material uses r$$_{21}$$ for F$$_{21}$$ which contradicts my previous attachment. Thats why i am confused.
Yeah, these things are always confusing. You always have to carefully look at how the author is defining everything.

As far as I can tell, each of your sources (EDIT: I mean attachements in this thread, not electric charges) uses a different definition for r21 and r12.

The first document is much more clear about the definitions, while the second one does not clearly define everything. However, the second reference is probably using one of the accepted conventions that r21 is equal to r2-r1, where r2 and r1 are the position vectors of the charges, using an arbitrary reference frame.

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studiot

Joined Nov 9, 2007
4,998
accepted conventions that r21 is equal to r2-r1
Remember this is vector subtraction!

logearav

Joined Aug 19, 2011
243
So r12 is r1 - r2? Am i right?

studiot

Joined Nov 9, 2007
4,998
I was not querying your statement, just reminding folks that r1 and r2 are vectors so the statement r1-r2 or r2-r1 is a vector subtraction.