Coulomb's law in vector form

1. Jan 4, 2012

logearav

1. The problem statement, all variables and given/known data

Revered members,

2. Relevant 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.

3. The attempt at a solution
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

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Last edited: Jan 4, 2012
2. Jan 4, 2012

cupid.callin

no attachment

3. Jan 4, 2012

logearav

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

4. Jan 5, 2012

ehild

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

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

5. Jan 6, 2012

ehild

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

ehild

6. Jan 7, 2012

logearav

Thanks for the help ehild.

7. Jan 7, 2012

ehild

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

ehild

8. Jan 10, 2012

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 ?

9. Jan 10, 2012

ehild

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