Question

What is the direction and magnitude of the force on the positive charge $+Q_0$ in terms of the given quantities?

• A. $\frac{-2{Q_0}^2}{4\pi {ϵ}_0{d}^2}\hat{i}$
• B. $\frac{-\sqrt{3}{Q_0}^2}{4\pi {ϵ}_0{d}^2}\hat{i}$
• C. $\frac{2{Q_0}^2}{4\pi {ϵ}_0{d}^2}\hat{i}$
• D. $\frac{\sqrt{3}{Q_0}^2}{4\pi {ϵ}_0{d}^2}\hat{i}$

Answer 1

The correct option is B $\frac{-\sqrt{3}{Q_0}^2}{4\pi {ϵ}_0{d}^2}\hat{i}$

As shown in figure the $\sin$ component of $F_1$ and $F_2$ cancels out each other and hence the resultant force is by $\cos$ component.,i.e.

$F_r=F_1\cos \theta +F_2\cos \theta$ $F_1=F_2\Rightarrow$Same distance and same magnitude.

$2F_1\cos \theta$

$\frac{2\times K\times \left({\theta }_o\right)(-{\theta }_o)}{d^2}\times \frac{\sqrt{3d}}{2d}$ $(\cos \theta =\frac{\sqrt{3}}{2}d-d)$

$=\frac{-\sqrt{3}K{\theta }_o^2}{d^2}$ [directed along $x$ axis)

$\frac{-\sqrt{3}{\theta }_o^2}{4\pi {ϵ}_o{d}^2}\hat{i}$