Coulomb's Law: Is It Correct to Use r12 Instead of r21 for Attractive Force?

[logearav]

Homework Statement

Revered members,
Please see my both attachments.

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

 

[cupid.callin]

Homework Statement

Revered members,
Please see my both attachments.

no attachment

 

[logearav]

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

 

[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

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

 

[ehild]

Correction:
Instead of

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

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

ehild

 

[logearav]

Thanks for the help ehild.
Btw, Both your quoted equation and the reply equation in your last reply was same.

 

[ehild]

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 ?

 

[ehild]

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