Day: December 19, 2023

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202312191041 Electrostatics Diagrams (Elementary) Q1

The blogger claims no originality of his problem below.

We have an arrangement of three positive (point) charges +Q_{A}, +Q_{B}, and +Q_{C} of the same quantity (=Q), as if lying in the vertices A, B, and C of an equilateral triangle \triangle ABC, where \mathbf{r}_{BA}, \mathbf{r}_{AC}, and \mathbf{r}_{CB} are the same magnitude r.

Intuitively, these three like charges are \textrm{\scriptsize{NOT}} in electrostatic equilibrium for acting between them are \textrm{\scriptsize{NET}} repulsive forces, not until we place a negative (point) charge -q of some yet unknown quantity q in the centre O, pointing to which some attractive forces of yet also unknown magnitudes.

Express q in terms of Q and r.

Setup.

\begin{aligned} |\mathbf{r}_{BA}| & = |\mathbf{r}_{AC}| = |\mathbf{r}_{CB}| = r \\ |\mathbf{r}'_{A}| & = |\mathbf{r}'_{B}| = |\mathbf{r}'_{C}| = \frac{\sqrt{3}}{3}r \\ \mathbf{r}_{BA} & = \mathbf{r}'_{A} - \mathbf{r}'_{B} \\ \mathbf{r}_{AC} & = \mathbf{r}'_{C} - \mathbf{r}'_{A} \\ \mathbf{r}_{CB} & = \mathbf{r}'_{B} - \mathbf{r}'_{C} \\ \text{}_{B}\mathbf{F}_{A} & = \text{}_{B}F_{A}\,\hat{\mathbf{r}}_{BA} \\ \text{}_{C}\mathbf{F}_{A} & = -\text{}_{C}F_{A}\,\hat{\mathbf{r}}_{AC} \\ \text{}_{C}\mathbf{F}_{B} & = \text{}_{C}F_{B}\,\hat{\mathbf{r}}_{CB} \\ \text{}_{A}\mathbf{F}_{B} & = -\text{}_{A}F_{B}\,\hat{\mathbf{r}}_{BA} \\ \text{}_{A}\mathbf{F}_{C} & = \text{}_{A}F_{C}\,\hat{\mathbf{r}}_{AC} \\ \text{}_{B}\mathbf{F}_{C} & = -\text{}_{B}F_{C}\,\hat{\mathbf{r}}_{CB} \\ \text{}_{B}F_{A} & = |\text{}_{B}\mathbf{F}_{A}| = F \\ \text{}_{C}F_{A} & = |\text{}_{C}\mathbf{F}_{A}| = F \\ \text{}_{C}F_{B} & = |\text{}_{C}\mathbf{F}_{B}| = F \\ \text{}_{A}F_{B} & = |\text{}_{A}\mathbf{F}_{B}| = F \\ \text{}_{A}F_{C} & = |\text{}_{A}\mathbf{F}_{C}| = F \\ \text{}_{B}F_{C} & = |\text{}_{B}\mathbf{F}_{C}| = F \\ \mathbf{F}_{A} & = \text{}_{B}\mathbf{F}_{A} + \text{}_{C}\mathbf{F}_{A} \\ \mathbf{F}_{B} & = \text{}_{C}\mathbf{F}_{B} + \text{}_{A}\mathbf{F}_{B} \\ \mathbf{F}_{C} & = \text{}_{A}\mathbf{F}_{C} + \text{}_{B}\mathbf{F}_{C} \\ \mathbf{F}_{A} & = F_{A}\,\hat{\mathbf{r}}'_{A} \\ \mathbf{F}_{B} & = F_{B}\,\hat{\mathbf{r}}'_{B} \\ \mathbf{F}_{C} & = F_{C}\,\hat{\mathbf{r}}'_{C} \\ F_{A} & = |\mathbf{F}_{A}| = F' \\ F_{B} & = |\mathbf{F}_{B}| = F' \\ F_{C} & = |\mathbf{F}_{C}| = F' \\ \end{aligned}

Write, by Coulomb’s law,

\begin{aligned} F & = k\frac{(Q)(Q)}{(r)^2} \\ & = k\frac{Q^2}{r^2} \\ \end{aligned}

such that

\begin{aligned} \text{}_{B}\mathbf{F}_{A} & = F\cos 60^\circ\,\hat{\mathbf{i}} + F\sin 60^\circ\,\hat{\mathbf{j}} \\ & = \bigg( k\frac{Q^2}{r^2}\bigg) \bigg(\frac{1}{2}\bigg) \,\hat{\mathbf{i}} + \bigg( k\frac{Q^2}{r^2}\bigg) \bigg(\frac{\sqrt{3}}{2}\bigg) \,\hat{\mathbf{j}} \\ \text{}_{C}\mathbf{F}_{A} & = -F\cos 60^\circ\,\hat{\mathbf{i}} + F\sin 60^\circ\,\hat{\mathbf{j}} \\ & = -\bigg( k\frac{Q^2}{r^2}\bigg) \bigg(\frac{1}{2}\bigg) \,\hat{\mathbf{i}} + \bigg( k\frac{Q^2}{r^2}\bigg) \bigg(\frac{\sqrt{3}}{2}\bigg) \,\hat{\mathbf{j}} \\ \mathbf{F}_{A} & = \text{}_{B}\mathbf{F}_{A} + \text{}_{C}\mathbf{F}_{A} \\ & = 0\,\hat{\mathbf{i}} + \bigg( k\frac{Q^2}{r^2}\bigg)(\sqrt{3})\,\hat{\mathbf{j}} \\ F' & = |\mathbf{F}_A| \\ & = \bigg( k\frac{Q^2}{r^2}\bigg)(\sqrt{3}) \\ \end{aligned}

and

\begin{aligned} F' & = k\frac{(Q)(q)}{(\sqrt{3}r/3)^2} \\ \bigg( k\frac{Q^2}{r^2}\bigg) (\sqrt{3}) & = \bigg( k\frac{Qq}{r^2}\bigg) (3)\\ q & = \frac{\sqrt{3}}{3} Q \\ \end{aligned}

\therefore We should put a negative charge -q of quantity \sqrt{3}Q/3 in order for the system to reach electrostatic equilibrium.

This problem is not to be attempted.