Question

Three electric charges $+q$ each are placed at the three corners of a square of side $d$. The intensity of electric field at the fourth corner is :

• A. $\frac{1}{4\pi {ϵ}_0}\frac{q}{d^2}(2+\frac{1}{\sqrt{2}})$
• B. $\frac{1}{4\pi {ϵ}_0}\frac{q}{d^2}(\sqrt{2}+\frac{1}{2})$
• C. $\frac{1}{4\pi {ϵ}_0}\frac{2Q}{d^2}$
• D. $\frac{1}{4\pi {ϵ}_0}\frac{\sqrt{2}q}{d^2}$

Answer 1

Answer: (B)

$\frac{1}{4\pi {ϵ}_0}\frac{q}{d^2}(\sqrt{2}+\frac{1}{2})$

Now E due to charge q at A is $\frac{kq}{d^2}(-\hat{y}).$
E due to charge q at C is $\frac{kq}{d^2}(-\hat{x})$
E due charge q at B is $\frac{kq}{(\sqrt{2}d{)}^2}cos{45}^0(-\hat{x})+\frac{kq}{(\sqrt{2}d{)}^2}sin{45}^0(-\hat{y})$
$=\frac{kq}{2\sqrt{2}{d}^2}(-\hat{x})+\frac{kq}{2\sqrt{2}{d}^2}(-\hat{y})$
Now total $E=\frac{kq}{d^2}(1+\frac{1}{2\sqrt{2}})(-\hat{x})+\frac{kq}{d^2}(1+\frac{1}{2\sqrt{2}})(-\hat{y})$
Magnitude of $E=\sqrt{E_{x^2}+E_{y^2}}$
$E=\frac{kq}{d^2}(\sqrt{2}+\frac{1}{2})$