# Coulomb's law in vector form

logearav

Revered members,

## Homework Equations

F21 = (q1q2/4∏ε0r122)*r12cap(unit vector)
Is it wrong to use r12 instead of r21 for F21. Because my second attachment uses r21 for F21 and r12 or F12. I am confused. Please help which is correct.

## The Attempt at a Solution

#### Attachments

• title.jpg
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• title1.jpg
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Last edited:

cupid.callin

## Homework Statement

Revered members,

no attachment

logearav
Sorry cupid.callin. Now i have incorporated the attachments.

Homework Helper
It depends how the forces and the unit vectors r12 and r21 were named. The left poster calls the unit vector pointing from 1 to 2 by $\hat{r}_{12}$, in the right one it is denoted by $\hat{r}_{21}$.

One is sure: the Coulomb force a charge exerts on an other charge acts in the line that connects them and repulsive when the charges are of the same sign.

If the position of two point charges are given with the vectors r1 and r2 then the force $\vec{F_{21}}$ exerted on charge 2 by charge 1 is

$$\vec F_{21}=k\frac{Q_1 Q_2}{(\vec {r_2}-\vec {r_1})^3}(\vec {r_2}-\vec {r_1})$$.

You can call the vector pointing from 1 to 2 by $\vec r_{12}$. The unit vector pointing from1 to 2 is

$$\hat r_{12}=\frac{\vec {r_2}-\vec {r_1}}{|\vec {r_2}-\vec {r_1}|}=\frac{\vec r_{12}}{r_{12}}$$

With this notation, the Coulomb force on charge 2 exerted by charge 1 is

$$\vec F_{21}=k\frac{Q_1 Q_2}{r_{12}^2}\hat r_{12}$$.

Homework Helper
Correction:
$$\vec F_{21}=k\frac{Q_1 Q_2}{(\vec {r_2}-\vec {r_1})^3}(\vec {r_2}-\vec {r_1})$$.

$$\vec F_{21}=k\frac{Q_1 Q_2}{|\vec {r_2}-\vec {r_1}|^3}(\vec {r_2}-\vec {r_1})$$

ehild

logearav
Thanks for the help ehild.

Homework Helper
I mistakenly used parentheses () in the quoted equation: It has to be magnitude instead ||.

ehild

logearav
Thanks again. If charges are opposite, then attractive force exists. Will the coulomb law take a negative sign? That is F = -KQ1Q2/r^2 ?

Homework Helper
Thanks again. If charges are opposite, then attractive force exists. Will the coulomb law take a negative sign? That is F = -KQ1Q2/r^2 ?

NO. One of the Q-s is negative, the other one is positive. Their product is negative so the force is negative. The law is the same for any Q1, Q2.

$$\vec F_{21}=k\frac{Q_1 Q_2}{r_{12}^2}\hat r_{12}$$

ehild