91Ó°ÊÓ

After reading the question, In the equilateral triangle with sides of $1.7\mathrm{m}$,

The electrical power is,

$\begin{array}{c}\mathrm{P}=\frac{\mathrm{dU}}{\mathrm{dt}}\\ =\text{0.83KW}\\ =0.83×{\text{10}}^3\text{W}\end{array}$

Potential Energy of a system of charges and relation between power, energy and time.

Formulae:

The potential is define by,

$\begin{array}{c}\mathrm{V}=\frac{1}{4{\mathrm{Ï€Î\mu }}_{\mathrm{o}}}â‹\dots \frac{\mathrm{q}}{\mathrm{R}}\\ =\mathrm{k}\frac{\mathrm{q}}{\mathrm{R}}\end{array}$

Here,${\mathrm{Î\mu }}_0$is the permittivity of free space, $\mathrm{q}$ is the electron charge, and $\mathrm{R}$ is the distance, and $\mathrm{k}$ is the Coulomb’s constant having a value$9×{10}^9{\text{Nm}}^2/{\text{C}}^2$.

The change in potential is,

$\mathrm{Δ\pm «}=({\mathrm{V}}_{\mathrm{f}}−{\mathrm{V}}_{\mathrm{i}})\mathrm{q}$

The time is defined by,

$\mathrm{t}=\frac{\mathrm{Δ\pm «}}{\mathrm{P}}$

From the figure:

Initial Potential at the tip of the triangle,

$\begin{array}{c}{\mathrm{V}}_{\mathrm{i}}=\frac{{\mathrm{kq}}_1}{\mathrm{r}}+\frac{{\mathrm{kq}}_2}{\mathrm{r}}\\ =\frac{\mathrm{k}}{\mathrm{r}}({\mathrm{q}}_1+{\mathrm{q}}_2)\\ =\frac{(9×{10}^9{\text{Nm}}^2/{\text{C}}^2)(12.0\text{C}+12.0\text{C})}{\left(1.7\text{m}\right)}\\ =1.27×{10}^9\text{V}\end{array}$

When 3rd charge bring to the midpoint between the 2 charges,

$\begin{array}{c}{\mathrm{V}}_{\mathrm{f}}=\frac{{\mathrm{kq}}_1}{\raisebox{1ex}{$\mathrm{r}$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}+\frac{{\mathrm{kq}}_2}{\raisebox{1ex}{$\mathrm{r}$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\\ =\frac{2\mathrm{k}}{\mathrm{r}}({\mathrm{q}}_1+{\mathrm{q}}_2)\\ =\frac{2×(9×{10}^9{\text{Nm}}^2/{\text{C}}^2)(12.0\text{C}+12.0\text{C})}{\left(1.7\text{m}\right)}\\ =2.54×{10}^9\text{V}\end{array}$

Therefore, the required energy is,

$\begin{array}{c}\mathrm{Δ\pm «}=({\mathrm{V}}_{\mathrm{f}}−{\mathrm{V}}_{\mathrm{i}})\mathrm{q}\\ =(2.54×{10}^9\text{V}−1.27×{10}^9\text{V})×12.0\text{C}\\ =(1.34×{10}^9\text{V})×12.0\text{C}\\ =1.5×{10}^8\text{J}\end{array}$

Thus the required time is,

$\begin{array}{c}\mathrm{t}=\frac{\mathrm{Δ\pm «}}{\mathrm{P}}\\ =\frac{1.5×{10}^8\text{J}}{0.83×{10}^3\text{W}}\\ =(1.807×{10}^5\text{s})\left(\frac{\text{1day}}{8.64×{10}^4\text{s}}\right)\\ =2.1\text{days}\end{array}$

Hence, the required time to move the point charges is $2.1\text{days}$.