Thank you for reporting, we will resolve it shortly
Q.
In the given figure, total work done $(W)$ by the external force from $r^{\prime}=\infty$ to $r^{\prime}=r$ is
Electrostatic Potential and Capacitance
Solution:
Electrostatic force between any two positive charges $q_{1}$ and $q_{2}$ separated by a distance $r$ is given by
$F _{21}=\frac{q_{1} q_{2}}{4 \pi \varepsilon_{0} r^{2}} \hat{ r }$ ...(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 $r=| r |$.
From Eq. (i), in this case $q_{1}=+Q$ and $q_{2}=+1 C$
$\Rightarrow \quad F =\frac{Q \times 1}{4 \pi \varepsilon_{0}\left(r^{\prime}\right)^{2}} \hat{ r }^{\prime}$ ...(ii)
Total work done $(W)$ by the external force is obtained by integrating Eq. (ii) from $r^{\prime}=\infty$ to $r^{\prime}=r$
$W=-\int\limits_{\infty}^{r} \frac{Q}{4 \pi \varepsilon_{0} r^{\prime}} d r^{\prime}=\left.\frac{Q}{4 \pi \varepsilon_{0} r^{\prime}}\right|_{\infty} ^{r}=\frac{Q}{4 \pi \varepsilon_{0} r}$
