Question

Here are the charges and coordinates of two charged particles located in an xyplane:${\mathbf{q}}_2^{\text{'}}\mathbf{=}\mathbf{-}\mathbf{3}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{\times }\mathbf{}{\mathbf{10}}^{-6}\mathbf{}\mathbf{C}$,$\mathbf{x}\mathbf{=}\mathbf{+}\mathbf{3}\mathbf{.}\mathbf{50}\mathbf{}\mathbf{cm}$,$\mathbf{y}\mathbf{=}\mathbf{+}\mathbf{0}\mathbf{.}\mathbf{500}\mathbf{}\mathbf{cm}$androle="math" localid="1662728679285" ${\mathbf{q}}_2^{}\mathbf{=}\mathbf{-}\mathbf{4}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{\times }\mathbf{}{\mathbf{10}}^{-6}\mathbf{}\mathbf{C}$,$\mathbf{x}\mathbf{=}\mathbf{-}\mathbf{2}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{cm}$,$\mathbf{y}\mathbf{}\mathbf{=}\mathbf{+}\mathbf{1}\mathbf{.}\mathbf{50}\mathbf{}\mathbf{cm}$. How much work must be done to locate these charges at their given positions, starting from infinite separation?

Short answer

The work done is$\mathrm{W}=-1.39\text{\hspace{0.17em}J}$

Step-by-step solution

Given

After reading the question, For the first charge ${\mathrm{q}}_1^{}=+3.0\times {10}^{-\text{6}}\text{\hspace{0.17em}C}$at $\mathrm{p}(3.5,0.5)\text{\hspace{0.17em}cm}$for the second charge ${}^{\text{'}}{\mathrm{q}}_2^{\text{'}}=-4.0\times {10}^{-6}\text{C}$ at $\mathrm{q}(-2,1.5)\text{\hspace{0.17em}cm}$
From the figure:

Determine the concept

Work done is given by s follows:

$\mathbf{W}\mathbf{=}\frac{1}{\mathbf{4}{\mathbf{\pi \epsilon }}_o}\cdot \frac{{\mathbf{q}}_1{\mathbf{q}}_2}{d}$

Calculate how much work must be done to locate these charges at their given positions

The separation between p and q,

$\mathrm{d}=\sqrt{{(3.5+2)}^2+{(0.5-1.5)}^2}$

$\mathrm{d}=5.59\text{\hspace{0.17em}cm}$

$\mathrm{d}=0.0559\text{\hspace{0.17em}m}$

The net work done:

$\mathrm{W}=\frac{1}{4{\mathrm{\pi \epsilon }}_{\mathrm{o}}}\cdot \frac{{\mathrm{q}}_1{\mathrm{q}}_2}{\mathrm{d}}$

Substitute the values and solve as:

$\Rightarrow \mathrm{W}=9.0\times {10}^9\times \frac{(3.0\times {10}^{-6})(-4\times {10}^{-6})}{\left(0.0559\right)}$

$\Rightarrow \mathrm{W}=-1.39\text{\hspace{0.17em}J}$