Three rigid rods are joined from an equilateral triangle ABC of side 1 m. Three particles carrying charges 20 $μ$ C each are attached to the vertices of the triangle. The whole system is at rest in an inertial frame.The resultant force on the charged particle at A has the magnitude

Zero

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3.6 N

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3.6

\sqrt{3}

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7.2 N

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Solution

The correct option is C 3.6 $\sqrt{3}$ N
$q_{1} = q_{2} = 20 μ C$
$r = 1 m$
$F_{A B} = F_{A C} = \frac{9 \times 10^{9} \times 20 \times 10^{- 6} \times 20 \times 10^{- 6}}{1}$
$= 36 \times 10^{- 1} = 3.6$
$F_{r e s u l t a n t} = \sqrt{F_{1}^{2} + F_{2}^{2} + 2 F_{1} F_{2} cos 60^{o}}$
$3.6 \sqrt{1 + 1 + 2 \times \frac{1}{2}}$
$= 3.6 \sqrt{3} N$

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Three rigid rods are joined from an equilateral triangle ABC of side 1 m. Three particles carrying charges 20μC each are attached to the vertices of the triangle. The whole system is at rest in an inertial frame.The resultant force on the charged particle at A has the magnitude

The correct option is C 3.6 √3 Nq1=q2=20μCr=1mFAB=FAC=9×109×20×10−6×20×10−61=36×10−1=3.6Fresultant=√F21+F22+2F1F2cos60o3.6√1+1+2×12=3.6√3N

Three rigid rods are joined from an equilateral triangle ABC of side 1 m. Three particles carrying charges 20 $μ$ C each are attached to the vertices of the triangle. The whole system is at rest in an inertial frame.The resultant force on the charged particle at A has the magnitude