Three $\mathbf{+}\mathbf{0}\mathbf{.12}\mathbf{\text{C}}$ charges form an equilateral triangle $\mathbf{\text{1.7m}}$on a side. Using energy supplied at the rate of $\mathbf{\text{0.83kW}}$, how many days would be required to move one of the charges to the midpoint of the line joining the other two charges?

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

The electrical power is,

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

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

Formulae:

The potential is define by,

$\begin{array}{c}\mathrm{V}=\frac{1}{4{\mathrm{\pi \epsilon }}_{\mathrm{o}}}\cdot \frac{\mathrm{q}}{\mathrm{R}}\\ =\mathrm{k}\frac{\mathrm{q}}{\mathrm{R}}\end{array}$

Here,${\mathrm{\epsilon }}_0$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$9\times {10}^9{\text{Nm}}^2/{\text{C}}^2$.

The change in potential is,

$\mathrm{\Delta U}=({\mathrm{V}}_{\mathrm{f}}-{\mathrm{V}}_{\mathrm{i}})\mathrm{q}$

The time is defined by,

$\mathrm{t}=\frac{\mathrm{\Delta U}}{\mathrm{P}}$

From the figure:

Initial Potential at the tip of the triangle,

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

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\times (9\times {10}^9{\text{Nm}}^2/{\text{C}}^2)(12.0\text{C}+12.0\text{C})}{\left(1.7\text{m}\right)}\\ =2.54\times {10}^9\text{V}\end{array}$

Therefore, the required energy is,

$\begin{array}{c}\mathrm{\Delta U}=({\mathrm{V}}_{\mathrm{f}}-{\mathrm{V}}_{\mathrm{i}})\mathrm{q}\\ =(2.54\times {10}^9\text{V}-1.27\times {10}^9\text{V})\times 12.0\text{C}\\ =(1.34\times {10}^9\text{V})\times 12.0\text{C}\\ =1.5\times {10}^8\text{J}\end{array}$

Thus the required time is,

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

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