Question

Three particles have charges + 20 $μ$ C each and they are fixed at the comers of a:r equilateral triangle of side 0.5 m. The force on each of the particles has magnitude

Easy

Solution

On every particle, there will be two forces of equal magnitude F and inclined at 60°. Here
$F = \frac{1}{4 {πε}_{0}} \times \frac{q_{1} q_{2}}{r^{2}} = (9 \times 10^{9}) \times \frac{(20 \times 10^{- 6}) (20 \times 10^{- 6})}{(0 \cdot 5)^{2}} \approx 14 \cdot 4 N$
Hence, resultant force is given by
$\begin{array}{l} = {[F^{2} + F^{2} + 2 FFcos 60]}^{1 / 2} = \sqrt{3} F \\ = 14 \cdot 4 \sqrt{3} N \end{array}$

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