Write down the Coulomb's law and get the vector form of it. - Vidyadip

Question

Write down the Coulomb's law and get the vector form of it.

✓

Answer

→ The electric force (Coulomb force) between two point stationary charges is proportional to the product of the values of the charges and inversely proportional to the square of the distance between them. The direction of this force is in the direction of the line joining the two charges.
→ Let the position vectors of charges $q_1$ and $q_2$ be $\vec{r}_1$ and $\vec{r}_2$ respectively [see Fig. (a)].

→ We denote force on $q_1$ due to $q_2$ by $\vec{F}_{12}$ and force on $q_2$ due to $q_1$ by $\overrightarrow{ F }_{21}$. The two point charges $q_1$ and $q_2$ have been numbered 1 and 2 for convenience and the vector leading from 1 to 2 is denoted by $\vec{r}_{21}$ :
$\vec{r}_{21}=\vec{r}_2-\vec{r}_1$

→ In the same way, the vector leading from 2 to 1 is denoted by $\vec{r}_{12}$ :
$\vec{r}_{12}=\vec{r}_1-\vec{r}_2=-\vec{r}_{21}$

→ The magnitude of the vectors $\vec{r}_{21}$ and $\vec{r}_{12}$ is denoted by $r_{21}$ and $r_{12}$, respectively $\left(r_{12}=r_{21}\right)$.
→ The direction of a vector is specified by a unit vector along the vector. To denote the direction from 1 to 2 (or from 2 to 1 ), we define the unit vectors:
$\hat{r}_{21}=\frac{\overrightarrow{r_{21}}}{r_{21}}, \hat{r}_{12}=\frac{\overrightarrow{r_{12}}}{r_{12}}, \hat{r}_{21}=-\hat{r}_{12}$

→ Coulomb's force law between two point charges

$q_1$ and $q_2$ located at $\vec{r}_1$ and $\vec{r}_2$, respectively is then expressed as
$\overrightarrow{ F }_{21}=\frac{1}{4 \pi \varepsilon_0} \frac{q_1 q_2}{r_{21}^2} \cdot \hat{r}_{21}$

→ Equation is valid for any sign of $q_1$ and $q_2$ whether positive or negative.
→ If $q_1$ and $q_2$ are of the same sign (either both positive or both negative), $F _{21}$ is along $\hat{r}_{21}$, which denotes repulsion, as it should be for like charges. If $q_1$ and $q_2$ are of opposite signs, $F_{21}$ is along $-\hat{r}_{21}=\hat{r}_{12}$, which denotes attraction, as expected for unlike charges.
→The force $\overrightarrow{ F }_{12}$ on charge $q_1$ due to charge $q_2$ is obtained from Eq. by simply interchanging 1 and 2 , i.e.,
$\overrightarrow{ F }_{12}=\frac{1}{4 \pi \varepsilon_0} \frac{q_1 q_2}{r_{12}^2} \hat{r}_{12}=-\overrightarrow{ F }_{21}$

→Thus, Coulomb's law agrees with Newton's third law.

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