- #1

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I'm quite accustomed with vectors (vector calculus and linear algebra background), but very new to e&m. Any help would be much appreciated.

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- Thread starter bifodus
- Start date

- #1

- 10

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I'm quite accustomed with vectors (vector calculus and linear algebra background), but very new to e&m. Any help would be much appreciated.

- #2

Tide

Science Advisor

Homework Helper

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The signs **are** included in Coulomb's law.

- #3

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I have a good book with an equation that should help you out.

Where the bolds are vectors, the regular font is scalar

and

where r1 is the location of the q1 and

r2 is the location of q2

If I'm not too clear, I'll fix it up some more

- #4

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[tex]\vec F_{on\ q_1\ due\ to\ q_2} = k\frac{q_1 q_2}{r_{12}{}^2} \hat r_{12}[/tex]

where [itex]\hat r_{12}[/itex] is the unit vector at the target charge [itex]q_1[/itex] pointing away from the source charge [itex]q_2[/itex] and

[itex]r_{12}[/itex] is the distance to the target charge [itex]q_1[/itex] from the source charge [itex]q_2[/itex].

If the product [itex]q_1q_2[/itex] is positive (so they have like signs), then,

since [itex]\hat r_{12}[/itex] points away from [itex]q_2[/itex], it follows that[itex]\vec F_{on\ q_1\ due\ to\ q_2}[/itex] points away from [itex]q_2[/itex].

"[itex]q_1[/itex] is repelled by [itex]q_2[/itex]."

If the product [itex]q_1q_2[/itex] is negative (so they have unlike signs), then,

since [itex]\hat r_{12}[/itex] points away from [itex]q_2[/itex], it follows that [itex]\vec F_{on\ q_1\ due\ to\ q_2}[/itex] points towards [itex]q_2[/itex].

"[itex]q_1[/itex] is attracted to [itex]q_2[/itex]."

- #5

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Thanks again.

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