ABC is a right angled triangle in which AB=3cm and BC=4cm and right angle is at B. The three charge,+15 $μ$ C,+12 $μ$ C and –20 $μ$ C are placed respectively at A,B and C, The force acting on B is

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2250N

3500 N

1200 N

1250 N

answer is D.

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Detailed Solution

$\begin{array}{l} F_{AB} = \frac{1}{4 {πε}_{0}} \frac{15 \times 10^{- 6} \times 12 \times 10^{- 6}}{{(3 \times 10^{2})}^{2}} \\ = \frac{9 \times 10^{9} \times 180 \times 10^{12}}{9 \times 10^{- 4}} = 1800 N = 900 \times 2 \\ F_{BC} = \frac{1}{({πε}_{0})} \frac{20 \times 10^{- 6} \times 12 \times 10^{- 6}}{{(4 \times 10^{- 2})}^{2}} \\ = \frac{9 \times 10^{9} \times 240 \times 10^{- 12}}{16 \times 10^{- 4}} = 1350 N \end{array}$
$\begin{array}{l} F_{net} = \sqrt{F_{AB}^{2} + F_{BC}^{2}} \\ = \sqrt{1800^{2} + 1350^{2}} \\ = \sqrt{5062500} = 2250 N \end{array}$
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