Question

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?

Short answer

The required time is $\mathrm{t}=2.1\text{days}$

Step-by-step solution

Step 1: Given data:

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

The electrical power is,

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

Determining the concept:

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,

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

Determining the energy:

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,

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

Determining the required days to move point charges:

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 $2.1\text{days}$.