Question

Two position vectors of point charges q₁ and q₂ are ${\overrightarrow{r}}_{1}$and ${\overrightarrow{r}}_{2}$, respectively. The electrostatic force of interaction between the charges is $F = \frac{q_{1}q_{2}}{4\pi\varepsilon_{0}r^{2}}$ where r is the distance between the charges and e₀ is the permittivity of free space.

The electrostatic force on first point charge q₁ due to second point charge q₂ in vector form is -

• A. $\frac{q_{1}q_{2}}{4\pi\varepsilon_{0}}\frac{({\overrightarrow{r}}_{1} - {\overrightarrow{r}}_{2})}{|{\overrightarrow{r}}_{1} - {\overrightarrow{r}}_{2}|^{3}}$
• B. $\frac{q_{1}q_{2}}{4\pi\varepsilon_{0}}\frac{({\overrightarrow{r}}_{2} - {\overrightarrow{r}}_{1})}{|{\overrightarrow{r}}_{2} - {\overrightarrow{r}}_{1}|^{3}}$
• C. $\frac{q_{1}q_{2}}{4\pi\varepsilon_{0}}\frac{{\overrightarrow{r}}_{1}}{|{\overrightarrow{r}}_{1} - {\overrightarrow{r}}_{1}|^{3}}$
• D. $\frac{q_{1}q_{2}}{4\pi\varepsilon_{0}}\frac{{\overrightarrow{r}}_{2}}{|{\overrightarrow{r}}_{1} - {\overrightarrow{r}}_{2}|^{3}}$

Answer 1

Answer: (A)

$\frac{q_{1}q_{2}}{4\pi\varepsilon_{0}}\frac{({\overrightarrow{r}}_{1} - {\overrightarrow{r}}_{2})}{|{\overrightarrow{r}}_{1} - {\overrightarrow{r}}_{2}|^{3}}$

Answer 2