Q31P - Introduction to Electrodynamics

Question

Three charges are situated at the comers of a square , as shown in Fig. 2.41. 

1. How much work does it take to bring in another charge, +q , from far away and place it in the fourth comer?
2.  How much work does it take to assemble the whole configuration of four charges?

                       

Step-by-Step Solution

Answer

$$\begin{gathered} \left(\mathrm{a}\right) \mathrm{The} \mathrm{work} \mathrm{it} \mathrm{take} \mathrm{to} \mathrm{bring} \mathrm{in} \mathrm{another} \mathrm{charge},+\mathrm{q} \mathrm{from} \mathrm{far} \mathrm{away} \mathrm{and} \mathrm{place} \mathrm{it} \mathrm{in} \mathrm{the} \\ \mathrm{fourth} \mathrm{comer} \mathrm{is} \mathrm{w}=\frac{{\mathrm{q}}^2}{4\mathrm{\pi }{\in }_0}\left\{\frac{1}{\sqrt{2}}-2\right\}. \\ \left(\mathrm{b}\right) \mathrm{The} \mathrm{work} \mathrm{does} \mathrm{it} \mathrm{take} \mathrm{to} \mathrm{assemble} \mathrm{the} \mathrm{whole} \mathrm{configuration} \mathrm{of} \mathrm{four} \mathrm{charges} \mathrm{is} \\ \mathrm{w}=\frac{1}{4\mathrm{\pi }{\in }_0}\frac{{\mathrm{q}}^2}{\mathrm{a}}\left\{\sqrt{2}-4\right\}. \end{gathered}$$

Step 1: Determine the expression for the potential.

Write the expression for the potential is,

$\mathbf{V}\mathbf{\left(}\mathbf{r}\mathbf{\right)}\mathbf{=}\frac{\mathbf{q}}{\mathbf{4}\mathbf{\pi }{\mathbf{\in }}_{\mathbf{0}}\mathbf{r}}$

Here, q is the charge, r  is the distance and  V is the volume.

Consider the figure for the charges.

                                                        

The figure consist the distribution of charged as shown above.

Consider the distance of the charge at point A form the charge +q is the diagonal of the square and the value is,

$$\begin{gathered} \mathrm{r}=\sqrt{{\mathrm{a}}^2+{\mathrm{a}}^2} \\ =\sqrt{2{\mathrm{a}}^2} \\ =\sqrt{2}\mathrm{a} \end{gathered}$$

Step 2: Determine the potential at point

(a)

 Now write the expression for potential at point A,

$$\begin{gathered} \mathrm{V}\left(\mathrm{A}\right)=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\left\{\frac{-\mathrm{q}}{\mathrm{a}}+\frac{\mathrm{q}}{\mathrm{a}\sqrt{2}}-\frac{\mathrm{q}}{\mathrm{a}}\right\} \\ \mathrm{Solving} \mathrm{for} \mathrm{V}\left(\mathrm{A}\right), \\ \mathrm{V}\left(\mathrm{A}\right)=\frac{1}{4{\mathrm{\pi \epsilon }}_0\mathrm{a}}\left\{\frac{1}{\sqrt{2}}-2\right\} \\ \mathrm{Thus},\mathrm{work} \mathrm{done} \mathrm{in} \mathrm{bringing} \mathrm{the} \mathrm{fourth} \mathrm{charge} +\mathrm{q} \mathrm{at} \mathrm{from} \mathrm{infinity}. \\ \mathrm{Let}\text{'}\mathrm{s} \mathrm{assume},\mathrm{V}(\infty )=0 \mathrm{is}, \\ \mathrm{w}=\mathrm{q}\left\{\mathrm{V}\left(\mathrm{r}\right)-\mathrm{V}\left(\infty \right)\right\} \\ \mathrm{Now} \mathrm{solving} \mathrm{for} \mathrm{V}\left(\mathrm{r}\right) \mathrm{by} \mathrm{substituting} \sqrt{2}\mathrm{a} \mathrm{for} \mathrm{r} \mathrm{in} \mathrm{equation} \mathrm{V}\left(\mathrm{r}\right)=\frac{\mathrm{q}}{4{\mathrm{\pi \epsilon }}_0\mathrm{r}}. \\ \mathrm{V}\left(\mathrm{r}\right)=\frac{\mathrm{q}}{4{\mathrm{\pi \epsilon }}_0\mathrm{a}\sqrt{2}} \\ \mathrm{Substitute} \mathrm{the} \infty \mathrm{for} \mathrm{the} \mathrm{r} \mathrm{in} \mathrm{the} \mathrm{equation} \mathrm{V}\left(\mathrm{r}\right)=\frac{\mathrm{q}}{4{\mathrm{\pi \epsilon }}_0\mathrm{r}}. \\ \mathrm{V}\left(\mathrm{r}\right)=\frac{\mathrm{q}}{4{\mathrm{\pi \epsilon }}_0\infty } \\ =0 \\ \mathrm{Substitute} 0 \mathrm{for} \mathrm{V}\left(\infty \right)\mathrm{and} \mathrm{V}\left(\mathrm{r}\right) \mathrm{in} \mathrm{the} \mathrm{equation} \mathrm{w}=\mathrm{q}\left\{\mathrm{v}\left(\mathrm{r}\right)-\mathrm{v}\left(\infty \right)\right\}. \\ \mathrm{w}=\frac{{\mathrm{q}}^2}{4{\mathrm{\pi \epsilon }}_0\mathrm{a}}\left\{\frac{1}{\sqrt{2}}-2\right\} \\ \mathrm{Therefore}, \mathrm{the} \mathrm{required} \mathrm{work} \mathrm{is} \mathrm{w}=\frac{{\mathrm{q}}^2}{4{\mathrm{\pi \epsilon }}_0\mathrm{a}}\left\{\frac{1}{\sqrt{2}}-2\right\}. \end{gathered}$$

Step 3: Determine the work required to assemble the whole configuration of the four charges.

(b)

 At point ,

Write the expression for amount of work done in bringing the charge,

$\mathrm{W}=\frac{\mathrm{q}}{4{\mathrm{\pi \epsilon }}_0}\left[-\mathrm{q}+\left(-\mathrm{q}+\frac{1}{\mathrm{a}\sqrt{2}}\right)+\left(-\mathrm{q}+\frac{1}{\mathrm{a}\sqrt{2}}-\mathrm{q}\right)\right]$

Thus, the work done in assembling the whole configuration is,

 $w=\frac{1}{4\pi {\epsilon }_0}\frac{q^2}{a}\left\{-1+\left(-1+\frac{1}{\sqrt{2}}\right)+\left(-q+\frac{1}{a\sqrt{2}}-1\right)\right\}$

Solve as further,

 $$\begin{gathered} \mathrm{w}=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{{\mathrm{q}}^2}{\mathrm{a}}\left\{\sqrt{2}-4\right\} \\ \mathrm{Therefore}, \mathrm{the} \mathrm{work} \mathrm{need} \mathrm{to} \mathrm{be} \mathrm{done} \mathrm{is} \mathrm{w}=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{{\mathrm{q}}^2}{\mathrm{a}}\left\{\sqrt{2}-4\right\}. \end{gathered}$$