Question

In the given figure, total work done $(W)$ by the external force from $r^{′}=∞$ to $r^{′}=r$ is

• A. $−\frac{Q}{4Ï€{Î\mu }_{0}r}$
• B. $\frac{Q}{4Ï€{Î\mu }_{0}r}$
• C. $\frac{2Q}{4Ï€{Î\mu }_{0}r}$
• D. $−\frac{2Q}{4Ï€{Î\mu }_{0}r}$

Answer 1

Answer: (B)

Electrostatic force between any two positive charges $q_1$ and $q_2$ separated by a distance $r$ is given by
$F_{21}=\frac{q_1{q}_2}{4Ï€{Î\mu }_0{r}^2}\hat{r}$ ...(i)

Here, $F_{21}$ is the electrostatic force on 2 due to 1 which along the vector $r$. Also, $r$ is the unit vector along $r$ and $r=âˆ\pounds râˆ\pounds$.
From Eq. (i), in this case $q_1=+Q$ and $q_2=+1C$
$⇒F=\frac{Q×1}{4Ï€{Î\mu }_0{\left(r^{′}\right)}^2}{\hat{r}}^{′}$ ...(ii)
Total work done $(W)$ by the external force is obtained by integrating Eq. (ii) from $r^{′}=∞$ to $r^{′}=r$
$W=−\underset{∞}{\overset{r}{∫}}\frac{Q}{4Ï€{Î\mu }_0{r}^{′}}d{r}^{′}={\frac{Q}{4Ï€{Î\mu }_0{r}^{′}}â}_{∞}^r=\frac{Q}{4Ï€{Î\mu }_{0}r}$