Q. Three charges 1 $\mu$C, 1$\mu$C and 2 $\mu$C are kept at vertices of $A,B$ and $C$ of an equilateral triangle $ABC$ of 10cm side respectively. The resultant force on the charge at $C$ is
A) 0.9 N
B) 1.8 N
C) 2.72 N
D) 3.12 N
Explanation
(d) [image:1_image312.emf]
$F_{AC} = \frac{1}{4\pi\varepsilon_{0}}\frac{q_{A}q_{c}}{(AC)^{2}}$
$= \frac{9 \times 10^{9} \times 1 \times 10^{- 6} \times 2 \times 10^{- 6}}{\left( 10 \times 10^{- 2} \right)^{2}} = 1.8$
Similarly,$F_{BC} = \frac{1}{4\pi\varepsilon_{0}}\frac{q_{B}q_{c}}{(BC)^{2}}$
$\ = \ \frac{9 \times 10^{9} \times 1 \times 10^{- 6} \times 2 \times 10^{- 6}}{\left( 10 \times 10^{- 2} \right)^{2}} = 1.8$
$F_{resultant} = \sqrt{F_{AC}^{2} + F_{BC}^{2} + 2F_{AC}F_{BC}\cos{60{^\circ}}}\ $
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \sqrt{(1.8)^{2} + (1.8)^{2} + 2(1.8)^{2} \times \frac{1}{2}}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 3.12\ $N
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.