Question

Three point charges $q_1={10}^{-6}C,q_2=-{10}^{-6}Cand{q}_3=0.5\times {10}^{-6}C$ are located in air at the corners of an equilateral triangle of 50 cm side, the magnitude of the force on $q_3$ will be ______ $\times {10}^{-3}$ N.

• A. 17.97

Answer 1

The correct option is A 17.97

The arrangement is shown in the figure. The triangle is placed in x-y plane with three corners at 0(0,0,0), Q(0.5,0,0) and P.

The distance PR can be obtained as,

$PR=\sqrt{d^2-{\left(\frac{d}{2}\right)}^2}=\frac{\sqrt{3}d}{2}$

$\therefore$ Co-ordinates of $P(\frac{d}{2},\frac{\sqrt{3}d}{2},0)$i.e. (0.25,0.433,0)

force on $q_3$ due to $q_1$ is

${\overrightarrow{F}}_{31}=\frac{q_3{q}_1}{4\pi {ϵ}_0{R}_{0p}^2}\overline{a_{0p}}$

where ,
$R_{0p}=0.25{\hat{a}}_x+0.43{\hat{a}}_y+0{\hat{a}}_z$

$\therefore$ ${\overrightarrow{F}}_{31}=\frac{0.5\times {10}^{-6}\times {10}^{-6}}{4\pi \times 8.854\times {10}^{-12}\times \left[\right(0.25{)}^2+\left(0.433{)}^2\right]}\frac{\overline{R_{0p}}}{|R_{0p}|}$

$=0.01805\frac{\left[\right(0.25\widehat{a_x+0.433\hat{a}y}+0{\hat{a}}_z\left)\right]}{0.5}$

$=8.99\times {10}^{-3}{\hat{a}}_x+0.01556{\hat{a}}_y+0{\hat{a}}_{z}N$

Force on $q_3$ due to $q_2$ is

$\overline{F_{32}}=\frac{q_3{q}_2}{4\pi {ϵ}_0{R}_{0p}^2}{\overline{a}}_{Qp}$

where ,
${\overline{R}}_{Qp}$ = $(0.25-0.5){\hat{a}}_x+(0.433-0)\widehat{a_y+0{\hat{a}}_{z}N}$
$\therefore$
$|\overline{R_{Qp}}|=-0.25{\hat{a}}_x+\left(0.433\right){\hat{a}}_y+0{\hat{a}}_z$
$|\overline{R_{Qp}}|$ = 0.5
$\therefore$
$\overline{F_{32}}=\frac{0.5\times {10}^{-6}\times (-{10}^{-6})}{4\pi \times 8.854\times {10}^{-12}\times (0.5{)}^2}\times \frac{[-0.25{\hat{a}}_x+0.433{\hat{a}}_y+0{\hat{a}}_z]}{0.5}$

$=0.00899{\hat{a}}_x-0.01556{\hat{a}}_y+0\widehat{a_{{}_z}}N$
$\therefore {\overline{F}}_3={\overline{F}}_{31}+{\overline{F}}_{32}$
$\therefore |{\overline{F}}_3|=0.01797N=17.97\times {10}^{-3}N$