Question

Consider the charges $q$, $q$ and $-q$ placed at the vertices of an equilateral triangle of each side $l$. The sum of forces acting on each charge is

• A. $\frac{q^2}{4\sqrt{2}\pi {\epsilon }_0{l}^2}$
• B. $\frac{-q^2}{4\pi {\epsilon }_0{l}^2}$
• C. $\frac{q^2}{4\pi {\epsilon }_0{l}^2}$
• D. zero

Answer 1

Answer: (D)

From diagram, force on $q_1(=q)$ at $A$,

${\vec{F}}_1={\vec{F}}_{12}+{\vec{F}}_{13}=F{\hat{r}}_1$
where $F=\frac{q^2}{4\pi {\epsilon }_0{l}^2}$ and ${\hat{r}}_1$, is the unit vector along $BC$
Force on $q_2\left(=q\right)$ at $B$, ${\vec{F}}_2={\vec{F}}_{21}+{\vec{F}}_{23}=F{\hat{r}}_2$
(where ${\hat{r}}_2$, is the unit vector along $AC$)
Force on $q_3(=-q)$ at $C$,
${\vec{F}}_3={\vec{F}}_{31}+{\vec{F}}_{32}=\left(\sqrt{F^2+F^2+2FFcos60^{\circ }}\right)\hat{n}=\sqrt{3}F\hat{n}$
where $\hat{n}=$ unit vector along the direction bisecting $\angle BCA$
$\therefore {\vec{F}}_1={\vec{F}}_2+{\vec{F}}_3=0$