Q. Charge $q_{1} = + \ 6.0$ nC is on Y- axis at y=+3 cm and charge $q_{2\ }$=-6.0 nC is on Y- axis at y=-3 cm calculate force on a test charge $q_{0} = 2$nC placed on X-axis at $x$=4 cm.
A) $- 51.8\ \widehat{j}\mu$N
B) $+ 51.8\ \widehat{j}\mu$N
C) $- 5.18\ \widehat{j}\mu$N
D) $5.18\ \widehat{j}\mu$N
Explanation
(a) Here, $q = \pm 6.0nC = \pm 6.0 \times 10^{- 9}C$
$2\ a = 6\ cm = 6 \times 10^{- 2}m$
$r = 4$ cm(on equatorial line)
$= 4 \times 10^{- 2}m$
$q_{0} = 2\ nC$
$= {2 \times 10}^{- 9}\ C,\ F = ?$
$F = F_{1}\cos{\theta + F_{2}\cos\theta}$
$= 2 \times \frac{1}{4\pi\varepsilon_{0}}\ \frac{qq_{0}}{r^{2}}\cos\theta$
[image:1_image363.emf]
$= 2 \times 9 \times 10^{9} \times \frac{6 \times 10^{- 9} \times 2 \times 10^{- 9}}{\left( 5 \times 10^{- 2} \right)^{2}} \times \frac{3}{5}$
$= 5.18 \times 10^{- 5}N$
$\overrightarrow{F} = - 51.8\widehat{\dot{j}}\mu N$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.