Question

Three point charges, which initially are infinitely far apart, are placed at the corners of an equilateral triangle with sides d. Two of the point charges are identical and have charge q. If zero net work is required to place the three charges at the corners of the triangle, what must the value of the third charge be?

Short answer

The value of the third charge must be $\frac{\mathbf{-}\mathbf{q}}{\mathbf{2}}$.

Step-by-step solution

Potential energy

Potential energy is given by:

$\mathit{U}\mathbf{=}\mathit{k}\mathbf{\sum }_{\mathbf{i}\mathbf{<}\mathbf{j}}\frac{{\mathbf{q}}_{\mathbf{i}}{\mathbf{q}}_{\mathbf{j}}}{{\mathbf{r}}_{\mathbf{i}\mathbf{j}}}$

Where k is a constant and the system is made of n charges as ${\mathbf{q}}_{\mathbf{1}}\mathbf{,}{\mathbf{q}}_{\mathbf{2}}\mathbf{,}{\mathbf{q}}_{\mathbf{3}}\mathbf{,}{\mathbf{q}}_{\mathbf{4}}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}{\mathbf{q}}_{\mathbf{n}}$,

and the distance between these charges are $$\begin{gathered} {\mathbf{r}}_{\mathbf{12}}\mathbf{(}\mathbf{distance}\mathbf{}\mathbf{between}\mathbf{\hspace{0.33em}}{\mathbf{r}}_{\mathbf{1}}\mathbf{\hspace{0.33em}}\mathbf{and}\mathbf{\hspace{0.33em}}{\mathbf{r}}_{\mathbf{2}}\mathbf{)}\mathbf{,}{\mathbf{r}}_{\mathbf{23}}\mathbf{(}\mathbf{distance}\mathbf{}\mathbf{between}\mathbf{\hspace{0.33em}}{\mathbf{r}}_{\mathbf{2}}\mathbf{\hspace{0.33em}}\mathbf{and}\mathbf{\hspace{0.33em}}{\mathbf{r}}_{\mathbf{3}}\mathbf{)}\mathbf{,} \\ {\mathbf{r}}_{\mathbf{34}}\mathbf{(}\mathbf{distance}\mathbf{}\mathbf{between}\mathbf{\hspace{0.33em}}{\mathbf{r}}_{\mathbf{3}}\mathbf{\hspace{0.33em}}\mathbf{and}\mathbf{\hspace{0.33em}}{\mathbf{r}}_{\mathbf{4}}\mathbf{)}\mathbf{,}{\mathbf{r}}_{\mathbf{45}}\mathbf{(}\mathbf{distance}\mathbf{}\mathbf{between}\mathbf{\hspace{0.33em}}{\mathbf{r}}_{\mathbf{4}}\mathbf{\hspace{0.33em}}\mathbf{and}\mathbf{\hspace{0.33em}}{\mathbf{r}}_{\mathbf{5}}\mathbf{)}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.} \end{gathered}$$.

Determine the third charge

The potential energy of the system is:

$U=k\sum _{\mathrm{i}<\mathrm{j}}\frac{{\mathrm{q}}_{\mathrm{i}}{\mathrm{q}}_{\mathrm{j}}}{{\mathrm{r}}_{\mathrm{ij}}}$

From the given statement, the diagram can be drawn as:

For three charge system, the above expression can be written as:

$U=k\left(\frac{q_1{q}_1}{r_{12}}+\frac{q_2{q}_3}{r_{23}}+\frac{q_1{q}_3}{r_{13}}\right)$

The charges ${\mathrm{q}}_1={\mathrm{q}}_2=\mathrm{q}\hspace{0.33em}\mathrm{and}\hspace{0.33em}{\mathrm{q}}_3={\mathrm{q}}_{\mathrm{c}}$and distances ${\mathrm{r}}_{12}={\mathrm{r}}_{23}={\mathrm{r}}_{13}=\mathrm{r}$.

Thus,

$$\begin{gathered} U=k\left(\frac{qq}{r}+\frac{q{q}_c}{r}+\frac{q{q}_c}{r}\right) \\ =k=\left(\frac{qq}{r}+\frac{2q{q}_c}{r}\right) \end{gathered}$$

The work done to create the system is stored as its potential energy. For work done to be zero:

$$\begin{gathered} W=0 \\ 0=U \\ k\left(\frac{qq}{r}+\frac{2q{q}_c}{r}\right)=0 \end{gathered}$$

Solving further as:

$$\begin{gathered} \left(\frac{qq}{r}+\frac{2q{q}_c}{r}\right)=0 \\ {q}^2+2q{q}_c=0 \\ {q}^2=-2q{q}_c \\ {q}_c=\frac{-q}{2} \end{gathered}$$

Therefore, the third charge is$\frac{\mathbf{-}\mathbf{q}}{\mathbf{2}}$