Question

Three equal charges each $+q$ are placed (fixed) on the corners of an equilateral triangle of side a. Then, the Coulomb force experienced by one charge due to the rest of the two is :

• A. $\frac{1}{4\pi {\epsilon }_0}\frac{q^2}{a^2}$
• B. $\frac{1}{4\pi {\epsilon }_0}\frac{2q^2}{a^2}$
• C. $\frac{1}{4\pi {\epsilon }_0}\frac{\sqrt{3}{q}^2}{a^2}$
• D. zero

Answer 1

The correct option is C $\frac{1}{4\pi {\epsilon }_0}\frac{\sqrt{3}{q}^2}{a^2}$

Generally coulomb force between two charges $q_1$ and $q_2$ is $\overrightarrow{F}=\frac{q_1{q}_2}{4\pi {ϵ}_0{r}^3}\overrightarrow{r}$ where $r$ is the distance between the charges.

From figure, $BC=\sqrt{a^2-\frac{a^2}{4}}=\frac{\sqrt{3}}{2}a$

Let $\overrightarrow{F_O}$ be the coulomb force on charge at O due to rest charges at $A(a,0)$ and $B(\frac{a}{2},\frac{\sqrt{3}}{2}a)$

thus, $\overrightarrow{F_O}=\frac{q^2}{4\pi {ϵ}_0{a}^3}\left(a\hat{i}\right)+\frac{q^2}{4\pi {ϵ}_0{a}^3}(\frac{a}{2}\hat{i}+\frac{\sqrt{3}a}{2}\hat{j})=\frac{q^2}{4\pi {ϵ}_0{a}^2}(\frac{3}{2}\hat{i}+\frac{\sqrt{3}}{2}\hat{j})$

$\therefore |\overrightarrow{F_O}|=\sqrt{\left(\frac{q^2}{4\pi {ϵ}_0{a}^2}{)}^2\right(\frac{9}{4}+\frac{3}{4})}=\frac{1}{4\pi {ϵ}_0}.\frac{\sqrt{3}{q}^2}{a^2}$