Question

Equal charges, $q$ each are placed at the vertices A and B of an equilateral triangle ABC of side $a$. The magnitude of electric intensity at the point C is:

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

Answer 1

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

Electric field due to A

$={\overrightarrow{E}}_A=\frac{1}{4\pi {\epsilon }_0}\frac{q}{a^2}\cos \theta \hat{i}+\frac{1}{4\pi {\epsilon }_0}\frac{q}{a^2}\sin \theta \hat{j}$

And $={\overrightarrow{E}}_B=\frac{1}{4\pi {\epsilon }_0}\frac{q}{a^2}\cos \theta \left(\widehat{-i}\right)+\frac{1}{4\pi {\epsilon }_0}\frac{q}{a^2}\sin \theta \hat{j}$

$\therefore \overrightarrow{E}={\overrightarrow{E}}_A+{\overrightarrow{E}}_B$

$=2\times \frac{1}{4\pi {\epsilon }_0}\frac{q}{a^2}\sin \theta \hat{j}$

As $\theta ={60}^o$

$\therefore |\overrightarrow{E}|=2\times \frac{1}{4\pi {\epsilon }_0}\frac{q}{a^2}\times \frac{\sqrt{3}}{2}$

$=\frac{\sqrt{3}q}{4\pi {\epsilon }_0{a}^2}$